Plane and Spherical - Forgotten Books

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T H E B EQU E S T O F

NATHANIEL I; B OWDITCH ,

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I

PLANE AND SPHERICAL TRIGONOMETRY .

PART I., containing Rules, Examples, and Problems. This volumecontainsthe principal Rules in Logarithms, and in Plane and SphericalTrigonometry. The Examples and Problems exceed 400, and the

answeror result is given under each . Sixth edition, 43. cloth .

II

PLANE AND SPHERICAL TRIGONOMETRY.

PART II., containing the principal Formulae, Proofs of Rules in Logarithms and Trigonometry, are. given in PART I. and a series of upwards of 300Analytical Examples and Exercises, with Hints and

Answers. Sixth edition , 48. cloth.

The two Parts complete in one volume, 7a. 64 . cloth.

III

PROBLEMS IN ASTRONOMY,SURVEYING, and

NAVIGATION,with their SOLUTIONS. Designed for facilitating

Self-instruction in the principles of Navigation and NauticalAstronomy. This volume contains the SOLUTIONS of\ all the Problemsgiven in Trigonometry, Part I. Second edition, 63. cloth.

IV

HAND-BOOK FOR THE STARS ;or Rules for finding the Names and Positions of all the Stars of theFirst and Second Magnitude. Third edition, 48. 6d. cloth.

V

NAVIGATION AND NAUTICAL ASTRONOMY.

PART I., containing Rules and Examples. 78. 6d. cloth .

VL

NAVIGATION AND NAUTICAL ASTRONOMY.

PART II., containing Investigations and Proofsof the principal Rules

The two Parts complete in one volume. 143. cloth .

LONDON : LONGMANS AND CO., PATERNOSTER ROW.

PLANE AND SPHERICAL

TRIGONOMETRY.

PART I.

CONTAINING ‘

BULES, EXAMPLES, AND PROBLEMS.

H . W. g iANS,LATE MATH EMATICAL MASTER

AND M INER AT THE ROYAL NAVAL COLLEGE, PORTSMOUTHPou l “ ! I ATB I I ATICAL EASTER IN TH E ROYAL I ILITABY ACADBKY, WOOLWICK

AI DAl RIAI INBB OP l l BCl ANT 07 3 103 8 8 I! NAUTICAL ABTBOI OH

“ DI R Tn m m BOARD.

DESIGNED PRINCIPALLY FOR NAVAL STUDENTS

AND ENGHVE’ERS.

Sixthchitin .

QLONDON

LONGMANS,GREEN

,CO . , PATERNOSTER ROW.

1 873.

DONDON

BORN )! ANDM RS, PR INTERS, PANCBAS ROAD

Tm; following pages contain the PracticalRules in Plane and

Spherical Trigonometry. The investigations of these Rules,requiring some knowledge ofMathematics, are ,

given in the

Second Part of this treatise. The present volume is madeto consist entirely of the Rules and their applications, and

the demonstrations are placed in Part II. , in order that the

student may be enabled to proceed with Problems in Surveying, Navigation, and Astronomy, as soon as he is acquainted

with vulgar and decimal fractions, and has acquired a sufli

cient knowledge of algebra to work an easy equation .

The Problems at the end of the book have been selectedchiefly forNaval Students ; most of them can be solved bymeans of the Rules contained in the book . The Problems

marked with'

the letter (a) have been added for the use of

those who have alreadymade some progress in Mathematicsthese cannot be solved by a simple application of the Rules ;

they will present, however, little difficulty to the student who

has made himself acquainted with the contents of Part II.

The table of log. haversines”contained in Inman’

sNau

found in Mendoza Rios, calculated to five places of decimals ; the

table in Norie, called log. rising, may be formed from it by the addition of the constant log.3 80103. The

‘

above table of Mendoza Rioswas re calculated by theAuthor of the present work, and carried to

vi PREFACE .

tical Tables, and now generally known to Naval Students,reduces considerably the labour of working out some of the

Problems in Navigation it may also be applied with,equal

advantage to the principal cases in Plane and Spherical Tri

gonometry. Rules have, accordingly, been now for the first

time adapted to this table.

* Other rules are also given for

the same cases suited to the Logarithmic Tables in more

general use, such as Hutton’

s,or those in Norie’

s orRiddle’

s

works on Navigation .

The young studentwho may use this volume as an INTRODUCTION TO NAVIGATION will not find it necessary to read, at

first,more than a certain portion of it. The Articlesmarked

in the table of contents with the letter (n) may be sufficientfor this purpose . The Examples under each of the Rules thus

selected should be worked out, with the exception of a few

of the more difficult. The Examples that may be omitted

at the first reading are separated from the others by a line.

Such student may also solve a few of the Problems at the

end of the book, such as [l ] to [39] to and, as

sis: places of decimals, and arranged in a more convenient form foruse ; it was first inserted in Dr. Inman ’

sNautical Tables, in 1837, un

der the name of the table of log. ha/vcrsines. Subsequently a similartablewas printed in awork on Navigation by Lieut. Raper, and called

by him log. sine aqua/re. The reason for adopting the two last-mentioned nameswill appear from considering the formula by means of

which the table was constructed from the common table of log. Sines,

namely :

sin }; i vers. A [Trig Part IL formula

Aor, log. sin. square

—

5=log. half versed sine A.

Rapcrderived the name of his table from the first side of this equa

tion Inman from the last, contracting half versed sine into haver

In the later editions of Inman ’s Tables a table of ha lf log. ha

versines has been added. This new table will be found to diminishthemechanical labourofworking outmany of the principalproblemsinNauticalAstronomy and Trigonometry.

PREFACE. vii

an INTRODUCTION TO NAUTICAL ASTRONOMY,Problems [102]

to The remainder of thebook may be omitted until he has read Part II.

The answers to theExamples and Problems have generallybeen taken from the tables by inspection ; that is, if Inman

’s

tables were at hand,to the nearest 15" if any others, to the

nearest minute or half minute. It is seldom necessary inpractical questions of this kind to proportion to the nearestsecond ; and in Nautical Problems the above degree of accu

racy will in almost every case be sufficient.Those who have not the assistance of a tutor, and who

have therefore to rely on their own exertions for the knowledge they may require, will find the volume of P roblems in

Astronomy, Surveying, and Navigation, with their Solutions,published separately

,and which in fact is the Key to the

problems at the end of this book, of great assistance.

H . W. J.

Langstone, near Havant,

Dec. 1872 .

CONTENTS.

PLANE TRIGONOMETRY.

The articlesmarked (0) may be omitted at first by the student. Those marked

(11) forman introduction toNavigation andNauticalAstronomy.

CHAPTER I.

(n) Trigonometrical ratios, viz. sine, tangent, &c.

(n) To find the index of the logarithm of any number.

(n) To find the logarithm of any number

( It) To find the natural number of a logarithm

( n) Multiplication by logarithms(n) Division by logarithms

( n) Involution by logarithms

(n) Evolution by logarithms .

(n) To find the value of anumberwith a fractional index( n) Proportion by logarithms

( it) To reduce algebraical formulae to logarithms .

(o) Exponential equations, 850.

(0) To find the value of an expression, when the index of

logarithm Is negative .

(7b) To take quantities out of table of logarithmic Sines, Sac.

( It) To take quantifies out of table of natural versines

( It) To reduce trigonometrical formulas to logarithms( n) To reduce logarithmic formulae to tabular logarithms(0) To calculate a natural versine from the table of logarithmic

sines, &c.

(0) Use of algebraical signs and to determine the magnitude

of angles

(0) To determine the algebraic sign of a trigonometrical ratio in

a formulawhere the other terms are known(0) To determine the numerical value of trigonometrical expres

sions connected by the signs and

CONTENTS.

APPLICATIONOF LOGARITHMS TO ARITHHETIO, MENSUBATION,&c. ; VIZ.

(o) Formulas for the mensuration of the principal regularbodies.

(0) Examples in mensuration

CHAPTER II.’

RULES IN PLANE TRIGONOMETRY.

(n) Rule I. Three sides of a plane triangle being given, to find

an angle (using table of haversines) . Firstmethod 44

(n) Three sides of a plane triangle being given, to find an angle

(without using Swondmethod

(n) Rule II. Of two sides and two opposite mgles, any threebeing given, to find the fourth

(o) Ambiguous case

(n) Rule III. Two sides and included angle being given, to find

the remaining angles

(n) Rule IV. Two sides and included angle being given, to find

the remaining side (using haversines) . Firstmethod

(n) Two sides and included angle being given, to find the remain

ing side (without haversines) . Secondmethod

(n) Rule V. Right angled plane triangles

(n) RuleVI. Two sides and included angle being given, to find

60

(a) RuleVII. Three sides of a plane trianglebeing given, tofind

the area

SPHERICAL TRIGONOMETRY.

CHAPTER III.

RULES IN SPHERICAL TRIGONOMETRY.

(n) Rule VIII. Three sides of a spherical triangle being given, to

find an angle (using haversines) . First method

(a) Three sides of a spherical triangle being given, to find an

(a) Rule IX. Two sides and included angle being given, to find

the third side (using haversines) . Firstmethod

CONTENTS.

(a) Two sides and included angle being given, to find the third

side (without haversines) . Second method

(n) Two sides and included angle being given, to find the third

side (using table of since, Thirdmethod

(n) Rule X. Of two sidesand two Oppositeangles, any three beinggiven, to find the fourth

(n) Rule XI. Two sides and included angle being given, to find

the other two angles

(n) Rule XII. Two angles and includedside being given, to findthe other two sides

(n) Rule XIII. Rules for right-angled spherical triangles( n) Rule XIV. Rules for quadrantal triangles

(n) ExAMINATION PAPERS IN PRECEDINC RULES

CHAPTER IV.

PROBLEMSm Suavmrme , kc.

PROBLEMS m SURVEYING, ASTRONOMY, ANDNAVIGATIONDEFINITIONSm ASTRONOMYDEFINITIONSINNAVIGATIONExAMPLEs AND PROBLEMS INNAUTICAL ASTRONOMY AND

VIGATION

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PLANE TRIGONOMETRY .

CHAPTER I.

(l ONE of the principal uses of Trigonometry is to furnish rules for finding any part of a triangle (that is, an angle

or a Side) when the number of parts already known is sufficient and this is the case

,in general, when three of the

parts of the triangle are known.

Considered asabranch ofmathematics, Trigonometryteaches us the method of investigating certain relations of

angles denominated formula . Trigonometry is therefore usually defined to be, that branch ofmathematics which treatsof the relations which the sides and angles of triangles haveto one another, and of the general relations of angles.

The First Part of this treatise contains the Rules for

solving Plane and Spherical h iangles, with a large collection

of examples and problems for practice. The problems havebeen selected principally to serve as an Introduction toNavi

gation and NauticalAstronomy.

In the Second Part will be investigated the funda

mental formula in Plans and Spherical Trigonometry. These

formula continually recur in mathematics, and the student

whose views extend beyond the practical part of Navigation

mustmake himselfwell acquaintedwith them. In the Second

Part also, the rules given in the First Part are investigated

and proved.

At present it will be sufficient to name only a few of

B

2 PLANE TRIGONOMETRY.

the terms used in Trigonometry the principal of which arethe fractions called the Trigonometrical ratios.

Let the angle PCNbe denoted byA

and let PN be drawn from any point P

perpendicular to ON, thus forming a

right—angled triangle then the trigono

metrical ratios are the fractions that can

be formed of the three sides of the right

angled triangle PCN, taking them two

and two together. It is evident thatonly six such fractions can be formed in thismanner, namely,PN CP PN CN CP CN

CP’

PN’

CN’PN’ DE ’

andCP

' To connect these frac

tions with one of the angles of the triangle, asA,let us call

the side CP opposite the right angle the hypothenuse, theside PN opposite to the angle in question the perpendicular,and the remaining side CN,

adjacent to the angle, the base ;then the trigonometrical ratios of an angle are named as follows the perpendiculardivided by the hypothenuse is calledthe SINE of the angle the hypothenuse divided by the per

pendicular is called the COSECANT of the angle the perpen

The use and importance of the trigonometrical ratios will be

more clearly seen in Part II. We may here briefly remark, that sincewe are not able to compare together the angles and sides ofatriangle

(for we cannot say that an angle is three ormore timcs greater thana side, since these quantities have difierent units of measurement) ,it hasbeen found convenient to take certain lines or fractions, such asPH

record these values in tables. Now these fractions will evidently depend for their value on the magnitude of the angles themselves ; andif we know the fraction, we shall know, by means of the table, the

angle corresponding to it. Moreover, we can compare the side of a

triangle with this fraction, since both quantities are expressed in thesame kind of measure, namely, linear measure ; and then, by reference to the table, discover the relation of the side to the angle itself.

6m , and compute their numerical values for difl erent angles,and

PLANE TRIGONOMETRY. 3

dieulet divided by the base is called the TANGENT of the angle

the base divided by the perpendicularis called the COTANGENTof the angle ; the hypothenuse divided by the base is called

the SECANT of the angle ; and the base divided by the hypothenuse is called the COSINE ofthe angle. An other term usedin Trigonometry is the VERSINE of an angle this quantity isequivalent to the excess of unity overthe cosine. These definitions will be more readily learned by writing them down inthe form of fractions, thus

CP hypothenuseIs called the sane ofthe angleA

hCP ypcosecant

tangent

cotangent

CP hyp .

CN basesecant

CN basecosine

CP hyp.

Also, 1 cosine ofA is called the versine ofangle A.

The values of these ratios orfractions being calculated foreveryminute or quarter ofaminute from 0° to constitutethe tables of since

,tangents, &c. , in general use : bymeans of

which,and certain other tables ofnumbers called logarithms,

problems in Trigonometry and Astronomy are usually solved.Before

,therefore, we proceed to apply Rules for solv

ing trigonometrical problems, we must, in order to familiarisethe student with logarithmic computation, Show the extensive

use.

of logarithms in shortening and simplifying complex and

tedious arithmetical calculations ; in fact, we shall find that,by theirmeans

,processes ofmultiplication are reduced to ad

dition,division to subtraction, the formation of powers to

multiplication, and the extraction of roots to division, &c.


In Part II. it is shown what is meant by a logarithm

,and how the logarithms of numbers have boon com

pated and arranged in tables, so that now the logarithm of

any given number may be found by inspection and con

versely, if any logarithm is given ,the number corresponding

to it (which, for the sake of distinction, is called the natural

number) may also be found.

LOGARITHMS.

’

EXAMPLES AND EXERCISES IN THE USE or LOCABITIIMS.

Logarithms usually consist ofawhole number and a deci

mal fraction : the whole numbermay be positive or negative ;the decimal part of the logarithm is always positive : thus,the student will find by referring to a table of logarithms thatthe logarithm of the natural number 200 is the

whole number, 2, is called the index or characteristic ; thedecimal part

,namely °301030, is called the mantissa. The

mantissaisalways positive ; the index maybe positive ornegative. Thus the logarithm is 2 and °301030;

but the logarithm is in fact 2 and

In the tables the decimal part or mantissa only is insertedthe index being found as follows

mixed number.

The index of the logarithm is less by one than the numberof integral places contamed in the number.

Thus the index ofthe log. of 24 is l

of 2463 is 3

of 147‘2 is 2

of is 1

of 14 72 is 0


RULE II.

To find the index of the logarithm of a decimal

Considerthe decimal fraction as awhole number, and find

the index of its logarithm by the preceding rule then sub

tract the number thus found from the number of figures in

the given decimal; the remainder, with the negative Sign

placed over it, is the index required .

Thus, the index of the logarithm of 00452 is 3 for the

index of the logarithm of 45 2 is 2,by Rule I. , and the num

ber of figures in 0045 2 is 5 : hence 2 — 5 — 3, or as it

must be written 3,to Show that the mantissa or decimal part

of the logarithm is still positive.

’

EXAMPLES.

The index of log. 1 23

of log.

‘0123

of log. 00064

of log.

°00000072l is

To find the index of the logarithm of a vulgar

Reduce the fraction to an equivalent decimal fraction ;the index of its logarithmmay then be found by Rule II.

Thus, the index of log.1 or of log. 1 25 is8

of log. or of log. is

lof log.

5-

5or of log. 04 Is

1

120

Hence, to reduce a logarithm with a negative index to a quan

tity wholly negative, and the converse, we must proceed thus

3 602060, or its equivalent 940; therefore, also,

or of log. 00833 is

6 PLANE TRICONOMETBY.

RULE III.

(1 To take out the logarithm of any given numberfrom

the table.

The rule for taking out a logarithm for any given num

ber will be found in the explanations which accompany the

table,to which the student is referred.

’ We will here insert

a collection of numbers for exercises whose logarithms it is

required tofind from the tables.

EXAMPLES.

Let it be required to take out from the tables the logar

ithms of 2482 , of and of °002482 .

Looking for 2482 in the table, we see opposite to it394802, the decimal part of its logarithm ; and this part, bythe explanation accompanying the table, is the same for eachof the above numbers : the index or whole number to beprefixed

,found by Rules I. orXL, will indicate the number

of integral figures in the natural number.

Thus the log. of 2482 is

log. of is 1 3 94802

log. of °002482 is

Find from the tables the logarithms of the following

natural numbers

24 Answer. Its logarithm is 1 -3802 1 1

24s 2 -39445 2

2480 3-394452

1476 31 69086

1 -147985

01 47985

3847

It might mislead the student to give in this place a rule for

taking out a logarithm, as it must be made to depend on the peculiar

arrangement of the table in his hands ; nor is it necessary, since

ample directions for finding the logarithms of numbers are always

appended to the table, either at the beginning or end.

8 PLANE TBIGONOME'I’BY.

tute the natural number, see Explanation accompanying thetables.

To find where the decimal point in the natural number

taken out must be p laced .

First. When the index of the given logarithm is positive,to find its natural number.

The number of integral places in the number taken out

will be one more than the index. Thus, if the index of a

logarithm is 2, the natural number contains three integral

places ; and so on : the rest of the figures in the numbertaken out are decimals.

EXAMPLES.

Given the logarithm 2 47712 1,to find its natural number.

Entering the table with the decimal part °477l2 1 , we

find the number corresponding to it to be 3, or 30, or 300, or3000

, 850 but,as the index ofthe logarithm is 2 , the natural

number must contain three integral figures. Hence the na

tural number of 7121 is 300.

Required the naturalnumbers ofthe following logarithms

No. is 24

248

2480

2480000

15 7098 3

1 °65 137

l21004 °l9


(l Second. When the index of the given logarithm is

negative, to find its natural number.

The natural number in this case will be a decimal. Tofind the numberofciphers (ifany) to be prefixed to the figures

taken from the tables, diminish the index by 1 (without t e

garding the negative Sign) ; the remainder will denote the

number of ciphers to be prefixed to the figures of the number

taken from the tables : thus,if the index is 4, prefix three

ciphers ; if 2 , prefix one cipher ; if 1 , the decimal mark is

placed next to the figures taken out and so on .

EXAMPLES.

Given the logarithm 445 2 , to find its natural number.

Entering the table with the decimal part°39445 2 , we

see the natural number opposite to it is 248 : to this number

prefix three ciphers,since the index of the logarithm is 4

hence the natural number of is 000248.

Required the natural numbers of the following logarithms

No. is

RULE V.

Multip lication by logarithms.

(1 Take out the logarithms of the given numbers fromthe tables add them together : their sum will be the logar

ithm of the product of the given numbers ; the naturalnumber corresponding to which is therefore the product required.

*

In Part II. it is shown that log. u mp m+log.p+ .


EXAMPLES.

Required the product of the following quantities

47x x 84 Ans.

°47x l40°5 x 0084°5 547

Calculation.

log. 47 °47

676

84 l °924279 0084

log. product 4405 3 log. product

product product

Required the product of the following quantities

1 . 72 x 96x 124 x05 Ans. 42854

2 84 x 96 8064

3 6x 4 x 12 x 32 9216

4 64 x 362 x 4 9267

5 . 36x 48x 62 x 4 42854

6. 1234 x 9671 x00617 73632

7. 2 4 x007x °54 x 1 0009072

8. 784 x000079 x °0000036 °0000002229

RULE VI.

Division by logarithms.

From the logarithm of the dividend subtract thelogarithm of the divisor: the remainderwill be the logarithmof the quotient ; the natural number corresponding to whichwill be the quotient required. *

EXAMPLE .

Divide 472 by Ans.

0472 by 01466

For (by Part log. log. n— log. m.

PLANE mroonomm r. l 1

log. 472 3942 log. 0472

log. 1 5 07856 log.

log. quotient 1 1 66086 log. quotient

quotient quotient

Find the value of the following expressions

9 . 68- 1— 34 Am.

10.

l l . -1- 34

12 . 19 - 72

13. 19

242 x 5 59 x 63781 x 43284 x 00769 x 6 83598x cocomox -039

116. l - 3 2 or 3125

12 -222

18. or 2187-23

RULE VII.

Involution, or raising ofpowers by logarithm .

(1 First. When the quantity to be raised to a poweris a whole ormixed number.

Multiply the logarithm of the quantity to be raised bythe number denoting the power ; and the product will be thelogarithm of the power, the natural number of which is thepower required.

EXAMPLES.

Required the l6th power of 105 , or the value of

For log. (n)'

r log. n (see Part IL) .

12 PLANE rmoorroun'rar.

log. 105 0021 189

16

127134

2 1 189

log.

‘339024

(1 2 18285

1 9 . Required the 6th powerof 47 2 15 Ans. 1 1078

20. 3d 125 195 3-127

2 1 . l5oth 105 1507-82

22 . 200th 10125 1 18 89

23. 4th 74 3701 53

24. 100001 1 0125 247742 3

Second. When the quantity to be raised to apower

is a decimal fraction .

Take from the tables the logarithm of the decimal frac

tion multiply,separately

,the decimal part of the logarithm

so taken out,and the index of the logarithm,

by the number

denoting the power.

From the latter product subtract (i. 8. add algebraically)the whole number in the former

,placing the negative sign

over the remainder and then affix to it the decimal part ofthe former product, the natural number of which will be therequired power which find from the tables.

EXAMPLES.

Required the l0th power of 2 ,or the value ofC2)

“

log. +°30103O

10 10

10 3010300

10

log, 7010300


Required the values of the following expressions

The 5th power of 2 or Ans. 00032°5 12

000070494

0063241

RULE VIII.

Evolution, or extracting of roots by logarithms.

First. Wh en the index ofthe logarithm ofthe givennumber is positive or if negative, divisible by the numberdenoting the root .

Take out the logarithm of the given number, and divideit by the figure denoting the root to be extracted ; and theresult will be the logarithm of the root

,which find from the

tables.

”

EXAMPLES.

Required the cube root of 1234,or f/ 1234.

Required the 5 th root of000052 14, or 3/000052 14.

Calculation.

[2]log.

-091315 log.

log. 471 234 1030438 log. <7000052 14

£7 1234 4 700005 214 1 391

29 . Required the 5th root of 784 Ans. 37 9195

30. square root of 365 191 0498

31 . cube root of 12345 231 1 162

32. l0th root of 2 1071776

33. square root of 093 °304959

34. 5 th root of 70825 9235

35 . 365th root of 1045 1000121

36. cube root of 00125 1 077

Second. When the index of the logarithm of the

log. nt For (Part II.) log. l/ n

T

14 PLANE rmooNonErEr.

number is negative, and is not divisible by the figure denot

ing the root.Take out the logarithm of the number whose root isre

quired ; increase the index by the least number of units that

will render it divisible by the figure denoting the root to be

extracted,annexing to the decimal part the same number of

units so added to the index : then proceed as before. Thus,to divide 4081241 by 3, write it down thus, 6then divided by 3 is 28 93747, the natural

number ofwhich is 078297.

EXAMPLES.

Required the square root of 1 45 2, or

Required the loth root of 00345 , or1

4700345 .

[1]

or

-1452 1 5 80983°381

Required the cube root of

square root of

72d root ofcube root of

RULE IX.

Tofind the value of a number raised to apowerrepresented

by afraction.

Multiply the logarithm of the numberby the nume

rator of the fraction, and divide the result by the denomin

ator ; the natural number corresponding to the quotient will

Required the value of (6-6, or

00345 3 537819

or 10

log.

1

4700345 5 3782

1700345 5 672

Am . 2 321

09644

06797

036342

16 PLANE TRIGONOMETEY.

EXAMPLES.

12 :

1079181

1098970

27 7815 1

077815 1

2000000

x= 100

Find the value of a: in the following proportions

48. 24 : 35 79 : :c Ans. x==1 15 °208

49 . 3505 : :c 1507 290 cc: 690

50. Find a fourth proportional to 0963,

'24958, and

008967 Ans. 02324

5 1 . 6027 : 56 1 596

RULE XI.

To reduce ari thmetical and algebraical expressions to log

arithms.

’

(a) The logarithm of the product of any number of

terms is equal to the sum of the logarithms of all the terms

in the product

Thus, if x=ab,then log. x=log. a+log. b

(b) The logarithm of the quotient of any two numbers is

equal to the logarithm of the dividend diminished by the

logarithm of the divisor

Thus, if cv=a — b,or 2

,then log. a:=1og. a— log. b

For the proof of this general rule see Part II.


abcde

log. ct r - log. a+log. b+ 1og. c— log. d— log. e

(c) The logarithm of the power of any quantity is equal

to the logarithm of the quantity multiplied by the number

denoting the power : thus, if x=a1°,then log. log. a.

If cc=a2b3, then log. w=2 log. a+ 3 log. 6.

(d) The logarithm of the root of any quantity is equal tothe logarithm of the quantity divided by the number denoting the root to be extracted

Thus, if then log. a:log

s

. a

If then log. as=§ log. a+zl; log. 6

or log. a

Reduce the following expressions to logarithms

5 2. z=abcd Ans. log. z=log. a+log. b+log. c+ log. d

53. a:a_o

blog. x=log. a+log. b log. 0

54. cc 371

1

)

log. r=log. u+log. b— log.,c— log. d

5 5 . sc=a2bcd2 log. z=2 log. u+log. b+log. c+ 2 log. d

2 aa

log. z=2 log. a+log. log. c— 1

.bJ a

d

./clog. log. a+log. b+i log. c— log. d

58. a: 23/

d log. log. (7+5 log. b— log.c 2 log. d

s 4

59. a:“if; u+§ log. b+§log. c—

7}log. d

Rules 7, 8, and 9 are derived from the same property

of logarithms, namely, log. a“

n log. a, where n may be anynumber, either whole or fractional : thus log. log. 6

log. C/7 log. 7 log. log. 2 .


If the index n of the quantity be negative, it will

be perhaps the easiest way to reduce it, in the first place, toan equivalent expression with a positive index, whichmay be

always done by substituting for the given quantity its reci

proca1,* with the sign of the power changed. For it is proved

by Algebra that

a =d"

1 1 1 r “ b?hence also

373-

7, 424 H a“

we may therefore easily find the logarithms of quantities

having negative indices by proceeding as follows : first, reduce

the given quantities to equivalent oneswith positive indices,and then turn these latter quantities into logarithms by thecommon rules already given : thus,

1

217Therefore log. a log. 1 n log. a

0 n log. a (since log. 1

dub— p

to logarithmsAgain, to reduce

n log. a+ g log. c—p log. b

similarly log. 27— 3=log. 3 log. 55 ; log. 4

— l=log

1 10 log. 4 ; log.

Hlog. log. a ; log.

p log. b— n log. a ; and so on .

EXAMPLES.

Required the value of the following expressions1 1_8

The reciprocal of any quantity as the reciprocal of a

whole numbern, orif 1

, and so on.

1 n

PLANE m eoNOMErar. 19

Calculation.

1 15

4— 5

4-2

— 7

0000000 log. 4 0002060 log. 2

2010300 5

log.

‘4— 5

. log.

13010300 3 204120

4— 5

°4— 5 970561

or7

log. 2 ? 1000589-2* °3987

Examples to the preceding rules.

the value of the following expressions

Ans.

04373

1 105910

“07509

00000675

1071 24

2187

1 1 6993

0004572

0548

0774

l007

1059

2 425


33x 45 x 1474.

35 x420022

75 . 32! and 150884 and 4098

78. and 85 538 and 7-103

77. (50) 100383

78. 5 4-98

79. (3124)los 2 1 1 -3

Sometimes examples of the following kind occur,in which several of the terms are connected by the signs

or In such cases the value of each term must be found

separately, and applied according to its sign, as in the following example

— 100x004

1 Calculation of numerator.

1076694 log.

0038347 log. 07 22931

6 07 781 5 1 log. 3

1016498 log.

and

572 log. 1087975

3

-272 188

(57 105 4437

2 . Calculation

log. 105 0021 189

20

log. 04 23780

log. 3 047712

log. 3 0000901

log. 13 50380

and log. (5 7 1054437

log. numerator=2 °404817

of denominator.

PLANE ratoONOME'rar. 21

3 =706 log. numerator 2 404817

and 100x004 '40 log. denominator 00785 22

denominator=7°56 log. fraction 1026295

and fraction=330 Ans.

Required the value of the following expressions

3 152

? 1

2Ans. 1747-8

6 a

145 37394 794

82. 33303

19*- 4-5 4

By means of the preceding rules the followingequations may be solved.

value of a: in the following equa

x=2 ‘658965

xs=l4 x=2 '4101

x=1 ’71323

x= l °3634

z3=004 ”1587

x4 =4%‘605 7

a: 06183

47691 x= 1005 4

x:

as:

a’ =b

0827

log. 6

log. a

log. c

log. a— log. b

log. c— log. b

m log. a— n log. 6


96. Ans. a:

99 . ab'=c

100.

101 .

RULE XII.

To find the value of a fractional logarithmic equation,

when the index of the logarithm of one or both of the terms

is negative.

Reduce the logarithm with the negative index to a quan

tity wholly negative (Rule II. p. The logarithm then

becomes wholly negative, and the division can be performed

by the common algebraical rule.

EXAMPLE.

Given 5 find the numerical value of 23.Since log. 04=log. 5

log. 5 0098970 0989733:

log.

‘04 2002060

w=

Find the numerical following equations

102 . 43“ 005 Ans. 2: 4

103. 001 a: 7 15

104. (04)cf=5 x=

105 . (04)

m log. b— log. a— rlog. 0

x= l7‘91

b log. a

log. c

1

log. b°

x= 1 °27

a: 0309

24 PLANE rmooNOMErEr.

tion it is seldom necessary to proportion to the nearest second,it being sufficiently correct to take out the quantity requiredby inspection . In Inman’

s Nautical Tables, the log. sine,&c.

ofan angle can be taken out by inspection to the nearestin nearly all otherNautical Tables to the nearest minute.

(28 When the angle whose trigonometrical ratio is re

quired is greater than its value may be found in the

table as follows : Subtract the angle from and look for

the remainder,which is called its supplement, in the tables.

Thus, to find the log. sine of 100° subtract it from

180° and look for the log. sine of the remainder (namely,79° which is 9093127 ; therefore log. sine 100

° 10'

9093127

But the readiest waywill, in general, be to diminish

the given angle by and to look out the remainder ao

cording to the following rule

RULE XIII.

To take outfrom the table the log. sine, log. cosine,

angle greater than

Diminish the angle by thenif sine is required look out cosine of remainder

cosine sine

tangent cotangent

and in the same manner for secant, cosecant, &c.

EXAMPLES.

108. Take out of the tables the log. cosine of 100°

the log. cosecant of 170° 14’

log. cos. 100°=log. sin . 10

°

log. cosecant 170° 14 ’

15 =log. sec. 80° 14’15

By the tableslog. sin . and

log. secant 80° 14' 15 107 70665

log. cos. and

log. cosecant 170° 14' 15

PLANE TRIGONOMETRY.

Table of natural versines.

EXAMPLES.

109. Find the natural versines of the following angles

26° 32

’1 5

” Ans. 105357

15 7 48 50 1925 964

90 7 15 1002109

125 0 30 1573695

1 175 443 Ans. 100° 6’16

105357 26 32 15

1925964 157 48 50

15 73695 125 0 30

There are other special tables used in Trigonometry and

Navigation, but they require no examples in this place toillustrate their use.

Reduction of trigonometrical formulas to tabular

Trigonometrical formula are mduced to logarithmsby the

common rules for reducing algebraical formula (22) but in

order to avoid,as much as possible

,the inconvenience of

using negative indices in calculations, the table of log. sines,

cosines, &c. is constructed by adding 10to the indices of thelog. sines,&c. Thus, the sine 30

° is proved in Trig. Part II. to

be equal to gor0,its logarithm must therefore be 1098970

but the log. sine of 30° contained in the tables, or as it iscalled the tabular log. sine, ,

is 9098970 that is, the tabularlog. sine is equal to the log. sine+ 10, and therefore the log.

sine of any angle=tab. log. sine— 10. And as the same ad

dition is made to the indices of the logs. of all the trigono

metrical ratios, we must, in reducing such trigonometricalformulae to tabular logarithms, subtract 10from the logarithm

of each. trigonometrical ratio used in the expressio n.


For a similar reason, viz. to avoid the use of negative in

dices, the table of natural versines is constructed by multi

plying the natural versines by one million thus, it is shownin Part II. that versine but the versine of 60

°

as contained in the table= °5 x that is,tab. vers.=ver. x 1000000; therefore, in logarithms

,

log. tab. vers.=log. vers. 6

and log. vers.=log. tab. vers.— 6

Hence we have the following rule for reducing trigonometrical

formulas to tabular logarithms.

RULE XIV.

To reduce a trigonometri calformula to tabular logarithms.

Subtract 6 hom each log. versine, and 10from the logar

ithm of each of the other trigonometrical terms that occur in

the formula.

EXAMPLES.

1 1 1 . Thus, if tan. A=sin. A. sec. A

In tabular logarithms,

tab. log. tan.A— 10=tab . log. sin .A 10+ tab . log. sec.A 10

or, as it is usually written ( suppressing .the word tabular, it

being understood that the formula is reduced to tabular lo

garithms),log. tan . A— 10=log. 81n A— 10+log. sec. A— 10

and collecting and cancelling the tens,log. tan . A=log. sin . A+ log. sec. A— 10

sin . AIf tan. A

cos. AIn tab. logs.

log. tan . A— 10=log. sin . A— lO— (log. cos. A— lo)or log. tan. A=1og. sin . A+ 10 - log. cos. A

1 12 . v er. z=2 sin . b sin . c sin .

log.ver.w — 10+2 log.sin

Aor log. vers. sin. b+log. sin. c+2 log. Sim-

2

PLANE mIGONOMErEr. 27

Reduce to tabular logarithms the following formula

1 13. sin. w=cosec. y tan. 2 1 14. a=b tan . A

1 15 . tan. A Z 1 18. cot. A

Answers to 1 13- 6.

1 13. log. sin. x=log. cosec. y-i- log. tan . z— 10

1 14. log. e=log. b+1og. tan . A— 10

1 15 . log. tan. A= 10+ log. a— log. b

1 16. log. w=log. u+log. cot. A— 10

In almost every problem in Trigonometry andNavigation

we shall have occasion to reduce trigonometrical formula to

tabular logarithms. We will therefore give in this place a

few more examples for reduction, and also for finding the

value of the unknown quantity byassigning numerical valuesto the given quantities. These examples will be found of

1 17. Given 3 cot. A; find the numerical value of a:

when a= 2o,and angle A=3O

°.

a:cot. A n=a cot. A

log. w=log. a+log. cot. A— lO.

sin.Acos. B1 18. Given tan s:

tan. CB=32° and O=7S°

Calculation.

log. a 1001030

log. cot.A 102 35648

1 1036678

10

1036678

x=34'41

tofindw,whenA=20°


tan. .1:em. A cos. B

In tabular logarithms,tan. C

log. tan. c — 10=log. sin . A— 10+log. cos. B— lo— (log. tan. C— lo)log. sin. A— 10+log. cos. B— lo— log. tan . C+lO

Cancelling the tens, we have

log. tan. w=log. sin. A+log. cos. B — log. tan . C

A=2O° 10' log. sin . A

B=32 45 cos. B 9024816

=78 45 194 62323

87 60985

z=3° 18'

1 19 . In the equation, s:=e tan. A sin . B cosec. C, find

33,having given a=416 feet, A=23° 50

’B=54° 28’

and C= 147° 32' Ans. x=278°7 feet.

RULE XV.

To find the natural versine of an angle by means of the

common table of logarithmic sines, &c.

Fira. When the angle is less than

Take out the log. cosine of the given angle and diminish

the index by 10, and find the natural number corresponding

sine of the angle . Subtract this natural cosine from un ity

the result will be the natural versine of the given angle,

which, multiplied by one million, will reduce it to the usual

Second. When the'

angle is greater than

Subtract 90° h'om the given angle. Take out the log. sine

of the result and diminish its index by 10 then find,as be

fore, the natural number of the remainder. Add unity tothenatural number, and the sum will be the natural versine of

the given angle. To get the usual tabular versine, multiply

the natural versine by 1000000.

PLANE rmooNOME'rEY. 29

120. EXAMPLE 1 . Find the tabular natural versine of

26° 32’

tabular log. cos. 28°

32'15 99 5 1850

10

log. cos. 26° 32'15 1 -95 1850

natural cos. 26° 32

'15

"08 94843

1 .

natural versme=0' 105357

tab. nat. versine=10535 7.

121 EXAMPLE 2. Find the tabular natural versine of

125°0

’

tabular log. sine 35° 0

’30

”

10

log. cos. 125°0

’30

”17 58681

natural cos. 125°

0'30 3695

1

natural versine=1 °573695

tab. nat. versine=1573695 .

Use of the algebraic signs and to determine

themagnitude of an angle.

The numerical value of any trigonometrical ratio of an

angle and of its supplement (28) being the same, therefore

the log. of any trigonometrical ratio taken out of the tablescorresponds to both angles: thus, sin . 100

° being equal tosin. the log. sine of 100

°=log. sine of For thisreason, when we have given the log. sine of an angle to find

the angle, we are not always certain that the angle found inthe tables or its supplement is the correct one. This uncertainty, however, can be removed when the trigonometrical

ratio in question is a tangent, cotangent, secant, or cosine,

by finding,by the following Rule, whether its algebraic sign

is or When the algebraic sign is the angle taken

out of the tables is the one sought when thealgebraic sign

30 PLANE ralGONOME'rar.

we must subtract the angle taken out from 180° for

RULE XVI.

To determine the algebraic sign of a trigonometri

cal ratio in anyformula adap ted to logarithms where all the

other terms are known, and thence to find the magnitude of

the angle, that is, whether it is greater or less than

Write down the formula, and simplify. it by clearing it of

fractions, &c. put over each given term its proper algebrai

cal sign ; that is, over each tangent, secant,cosine, and cc

tangent of an angle or arc less than the sign and

over each of the same quantities when the angle is greater

than the sign but over each sign, cosecant, and ver

sine, the sign whether the angle is greater or less than

and determine from thence the sign of the product of

that side whose terms are all known .

Then,

‘

since the sign of the product of each side of the

equation must be the same (otherwise we should have a po

sitivo quantity equal to a negative quantity), make it so

by putting over the unknown term the sign or accord

Then, if falls over the unknown term, the part required

is less than and the quantity taken outwill be the angle

required ; but if subtract the angle taken out of the table

from 1 the remainderwill then be the angle required.

(Ambiguous case) When the part sought is expressed

in terms of the sine, the above rule will not apply, sinco the

sine is positive,whether the angle is greater or less than

The uncertainty which thence arises (forming what is calledin Trigonometry the ambiguous case) can be removed only inparticular cases.

In the above rule the angle required is supposed to be less

thanEXAMPLES.

In each of the following formula it is required tofind the

value of a: supposingA=45°

,B=120°, and C=130

°.


124. Given tan . x= cosec. A cos. B .

Find whether a: is greater or less than

Placing over cosec. A and cos. B their proper algebraic

signs, we see that tan . a: is negative, or

4.

tan . x=cosec. A. cos. B

and therefore a: is greater than

125 . cosec. A Sin . B=oot. cc

Proceeding in a similar manner, we find that cot. a: is posi

tive, and therefore :c is less than

126. sec. A sin .

2C=cos.

2B cot. w.

Placing over the given quantities their proper Signs, and re

collecting that the square of cos. B . must be positive, we have

sec. A sin 20=cos.

2B cot. a:

a: is less than 90°

127. sec. A sin .

2C — cos.

2B cot. a:

Placing the Signs over the given quantities, and taking intoconsideration the negative Sign in front of the right-hand sideof the equation, we see that must be placed overcot. a: in

order to render the right-hand Side of the equation the same

Sign as the left, namely, positive hence

sec. A — cos.

2B cot. x

a: is greater than 90°

Sin. A cos. a: cos. B cot. C

w is greater than 90°

cos. B

Sin . C tan . .n

or ver. A Sin . C tan . x=cos. B

ver. A

ora: is greater than

PLANE TBIGONOMETBY. 33

Sometimes trigonometricalexpressions occurinwhich

several of the terms are connected by the signs or and

which cannot be reduced to a convenient logarithmic form.

To determine the numerical value of such expressions, the

value of each term must be found separately, and then ap

plied with its proper Sign . The method of proceeding will

be seen in the following example

130. Given u : .2/tan a: sec. y— aQb cos fi

y ; to find the

value of u , when w=32°

y=80°

,a=25 , and b==50.

Ans. u

Let m= tan . a: 800. y. n=a2b cosfiy

log. m=log. tan. a+ log. sec. y— 20

log. n 2 log. a+ log. b+ 2 log. cos. y— 20

and log. u=§ log. (m— n)

To find u .

log. log. (m— n)log. (m— n) 20725 17

4 log (m 0090839

u

0 0 0 0 0 0

Find u in the following expressions : the values of a, b, w,

and y, being the same as in the last example.

131 . n=a cos. at — b Sin. y. Ans. u — 2808

a sec. z+ b cosfiy

tan . zc— cos. y

Logarithms were originally invented to facilitate trigo

n

132 . u

To find n.

1097940

1097940

1098970

log. cos. y . . 92 39670

log. cos. y . . 9 2 39670

log. n 2 974190

942 3

302

— 938-88


nometrical calculations. The following examples will Show

their use in problems of another kind.

App lication of logarithms to questions inprogression,interest

,mensuration, &c.

IN GEOMETRICAL PROGRESSION. If e=the first term of

a geometrical series, r=common ratio, n=number of terms,and S=Sum of terms ; then it is proved in Algebra that

S=a Any three of these quantities being given,

the fourth may be readily found by means of logarithms.

EXAMPLES.

133. Find the sum of 20terms of the series.

r" (3

3

,20 1 1

r— l g— l Qlog. (1 log.

3325 2 1 — 1 33242 1

2 l134. The sum of a geometrical series is 65 60

, its first

term 2, and common ratio 3 find the number of terms.

and S Ans.

r" — 1 3fl — 1

r— l3 — 1

_ log. 65613 _ 6561 , and n log. 3— log. 6561 n

log. 3

135 . If sum of series= 1023,first term= l

,common ratio

— 2 what is the number of terms? Ans. n= 10.

Without the use oflogarithms the last two questionswould

be very difficult to answer.

IN COMPOUND INTEREST. If P denote the principal, r theinterest of £ 1 for 1 year, n the number of years, and A the

amount, then it is proved in Algebra that A=P (1 r)“

any three of these quantities being given, the fourth may be

found.


EXAMPLES .

136. If £ 200be placed out at compound interest for 7years at 4 per cent required the amount.

log. A=1og. P+n log. (1 log.

Ans. A=£ 263 38. 8d.

137. At what rate of interest must £ 400be placed outthat it may amount to £ 569 68. 8d. in nine years, at com

pound interest ?

A=P(1 r)” log. (1 + r)

Ans. r=04, or 4 per cent.

138. In how many years will £ 500amount to £ 900at 5

per cent, compound interest ?

A=P(l r)”

Ans. u=1204 years.

139 . IN ASTRONOMY. It is proved that the aqua-se of

the times which two planets employ to make their revolu

tions about the Sun are to each other as the cubes of their

distances from the same heavenly body. Knowing that the

revolution of the Earth about the Sun is performed in 365

days, 5 hours, 48 minutes, 5 1 seconds ; and that ofJupiter in4330days

,14 hours, 39 minutes, 2 seconds : it isrequired to

find the ratio of the distances of these planets from the Sun .

Reducing in the first place into seconds the given re

volutions, we have 31 5 56931 seconds for the Earth, and

374164742 for Jupiter. Representing the first by a,the se

cond by b, the distance of the Earth from the Sun by 1 , andthat of Jupiter by as, we have

23:

3 log. r= 2 log. b -2 log. a

z 1°g‘ 6

32 1°g‘ a

— O-715978 x=5 °l997

Or the distance of Jupiter from the Sun is to that of the

Earth from the same body nearly as 5 2 10.

and log. a:


Formula for theprincipal regular bodies.

circumference of circlet where w=3°1416 nearly.

Area ofrectangle=a b. a, b, adjacent sides.

parallelogram=abSin . C a, 5, Sides, 0included angle

triangle : J s . s— a s— b s— c a, b, c, Sides, and

triangle a 5 Sin . C a, 6, sides, C included angle

180°

regular polygon=4 n a2cot.

nn Sides each=a

circle==1rr2=zd2 d=diameter

annulus (dfl

(cl— do

exterior and interior diameters d and d1

ellipse=£ddl major and minor diameters d and d1

Surface (convex) of cylinder= 7rd h dia. d height h.

cone gdh1 die. (I slant height hl

parallelogramofsame

height and base, coordinates x, ySphere=¢d2=cd dia. d circumf. c

Volume of rectangular parallelop1ped=a bc a, b, c, sides

czh

2cylinder=—Id3h 079577 c h

dia. d height h circumf. c

cone g-

gdih= o2818d2h=0285 28ci h

dia. d height h circumf. c

sphere M = °5 236d3 dia. d.

prolate spheroid=§ x circumscribing cylinderor a b? a semi-major, b semi-minor axis

Volume of oblate spheroid=§ x circumscribing cylindera},era2 b a semi-major, b semi-minor axis

paraboloid=§ x circumscribing Cylinder

dia. d height h

PLANE raiSONOME'rEr. 37

Mensuration of areas, surfaces, and solids.

The following examples in mensuration are given chieflyas exercises in logarithms they can all be easily calculated

by means of the formula contained in page 36.

EXAMPLES.

140. Two adjacent Sides of a rectangle are 40} feet and281 feet : required its area.

area=a

141 . Two adjacent Sides of a rectangular field are 6750yards and 4730 yards : requiled its area or superficial content .

Ans. 31971206 sq. yards.

142 . The three sides a, b, c, of a plane triangle ABC are

a= 1 150, b=937°5,and c=687°5 yards : required itsarea or

superficial content.

b=937°5 log. area — c +log.s - b

+ log. s— c}

1387°5=s

237°5=s— a

450°0=s— b

— c

area

b (p.

area=log. a+ log b

a 100745 5

3088093

1 184 -375 sq.

31 42233

2075664

205 3213

2045098

5008104

area=322185 sq. yds.


143. The three Sides of a triangular field are. a=3762 ,

b=450°25 , c=5022 5 required its superficial content.

Ans. 815090

144. Two Sides a and b of a plane triangleABC are

5 22 5 yards and b=60°1 25 yards, and the included angle0

72° 32'

required its area.

Area=4 ab Sin. C. (p .

a=5 2 °25 log. area=log. a+ log. b+log. Sin . C— lO— log. 2

b=60°l25 log. a

C=72° 32’log. b

log. sin. C 9079499

10001030

log. area

area=1498°34 sq. yards.

145 . Two sides of a plane triangle ABC are b=812 °l25 ,

e= l015 2 5 , and angle A=57° 35

’

ze find its superficial con

tent.

Ans. 3480140

From the two formula used above are deduced Rules

VI. and VII. in Plane Trigonometry given hereafter for find

ing the area of a triangle.

146. A Side of a regular heptagon is yards : re

area=fn a?

cot. (p.

a=19 °38 25° 42'5 1

log. area=log. n+ 2 log. u+ log. cot 10— log. 4

40 PLANE m oouom ar.

148. The diameter of t circle is 111;yards : required its

surface or area.

d='

l {I area=ji'

d2

°785 4 x

log. area=log.

°785 4+ log. 49 — log.

109509 1

1090196

1 °585287

105 6303

0028984

area=1 °06901 sq. yards.

149. The equatorial diameter of the earth is about 7924

miles : what is the area of the terrestrial equator?

Ans. By logarithms 49315000miles

more accurately by arithmetic 49315090

Netc. The Table of Logarithms in general use being only

approximately correct must not be depended upon to give

correct results when the natural numbers concerned exceed

four or five figures this is seen in the above example.

150. The outer and inner diameters of a flat ring are re

spectively 40} and 304 inches required the area or super

ficial contents.

d =404=40°25

area

d+d,

d— dl

10125

W

(d — d,) (p.

1095091

1047419

1005 904

2 747904

surface of ring=559033 sq. in.


Additional examp les of the use of theformula in page 36

for calculating areas, surfaces, and solids.

1 5 1 . Required the area or flat surface of a circular ringwhose outer and inner diameters are respectively 204 and

1 5113 inches.

Ans. 139008 sq. in .

152 . The major and minor diameters of an ellipse are

respectively 1404} and 85§ inches : required the area or sur

face.

Ans. sq. in .

153. What is the superficial content of an ellipse whosediameters are 625 and 5620 yards?

Ans. 2761 17 sq. yards.

15 4 . The diameter of a cylinder is 20§ inches, and its

height is 5 5 inches required the convex surface and also thewhole surface of solid.

Ans. Convex surface=24 °9 sq. ft.

whole surface=290

15 5 . The diameter of a cylinder is inches,and its

length 5 2} inches : required the whole surface of solid.

Ans . sq. ft.

156. What is the whole surface of a cone whose slantside is 20feet and the diameter of the base 20647 feet ?

Ans. ft.

157. What is the whole surface of a cone whose slantSide is 20feet and the circumference of the base 9 feet ?

Ans. ft.

1 58. What is the surface of a Sphere whose diameter is

24 feet ?Ans. 180906 sq. ft.

159. The equatorial diameter of the earth is about 7924

miles ; what is the superficial content of a sphere of that

magnitudeAns. By logarithms 197260000miles ;

more accurately by arithmetic 197260360

42 PLANE TEISONOMETEY.

160. The length of a beam is 254 feet, breadth 2 feet6 inches

,and depth 1 foot 8 inches required its cubic con

tents.

a=25§=250b

c=1a= 13=i

161 . A beam 40feet Ring is

required its cubic contents.

Ans, cubic ft.

162 . The cubic contents of a piece of square timber is

1062 5 feet, the breadth 24 feet, and depth 1§ feet find the

length.

Ans. 250 ft.

163. The surface of a’

pond measures 10acres : what is

the weight of the water .withdrawn when the surface falls

14 inches by evaporation, supposing 1 cubic foot of water to

weigh 1000ounces

Ans. 15 19 tons nearly.

164. What is the capacity of a cylinderwhose height and

the circumference of its base are each 20. feet ?

Ans . 636018 cubic ft.

165 . Three cubic feet are to be cut off a cylinder 44

inches in circumference : what distance from the end must

the Section be made ?Ans. 3304 in .

A cubic foot of pure water at 39° F. (when its density is the

greatest) weighs 9980 ounces. An imperial gallon contains 2772 74

cubic inches. The imperial pound avoirdupoise is defined to be theweight of one-tenth of an imperial gallon of distilled water at the

temperature of 62° F. and when the barometer stands at 30inches

consequently the weight of an imperial gallon is 10lbs.

V=abc=25 °5 x 20x9,log. 250

20 0097940

5 0098970

2003450

3

V 2026329

V=1062 5 cubic feet.

inches square at the end :


166. The spire of a church of a conical form measures

370992 feet round its base its perpendicular height is 100

feet : required its cubic contents.

Ans. 3770cubic ft.

167. Required the cubic contents of a cone whose diameter is 7924 miles and perpendicular height 3962 miles.

Ans. by logarithms, 65 1 28600000miles

more correctly by arithmetic, 65 1 28795619

168. Required the volume of a globe whose diameter is30feet. Ans. 141372 cubic ft.

169 . Required the volume of a sphere whose diameter is79572miles.

Ans. by logarithms, 263857900000milesmore correctly by arithmetic, 26385 7437760

170. The axes, or diameters, of a prolate Spheroid are 40and 50: required its volume. Ans. 41888.

Note. A prolate spheroid is a solid generated by the revolution of an ellipse about its major diameter. An oblatespheroid is a solid formed by the revolution of an ellipse

about its minor diameter.

171 . What is the solidity of an oblate spheroid whose

axes are 201» and Ans. 21730.

172 . The earth is an oblate spheroid whose equatorialand polar diameters are respectively 7924 and 7900milesrequired its volume.

Ans. by logarithms,259726000000miles

more correctly by arithmetic, 25 9726141037

CHAPTER II.

EULESm PLANE TRIGONOMETBY.

RULE L— Fiasr MErnon, USING HAVEESINES.*

Three sides of a p lane triangle being given, to find an

angle.

Put down the two Sides containing the required angle,

and take the difference,under which put the third Side ;

take the sum and difference,and also the half sum and half

To the arithmetical complementst of the logarithms of

the first two terms in this form add the logarithms of the

last two, and reject 10 in the index ; the result will be'

thelog. haversine of the required angle, which find in the table.

This rule is derived from formulaB (Part II. )

be hav. A=§ (a+c ) ii (a— bcv c) .

The mathematical student will no doubt in some cases

preferti) work out the examplesby the aid ofaformularather

than by the rule deduced from it. The formula for solvingtriangles will be found investigated in Part II.

If the student have no table of log. haversines, the anglemaybe found by the second method (p.

1° The arithmetical complement of a logarithm is the dif erencc

between the logarithm and 10; thus lo— log . 3 is the arithmetical

complement of log. 3. In practice it is most easily found by takingeach figure of the logarithm from 9, except the last, and that from10 ; thus,ar. cc . of 2 714152 is 72 85848 ; ar. co. 1014150is 10085850

(the last figure in this Example being 5 , the cipher at the end not

being of any value in that position) .

PLANE TRIGONOMErEr. 45

EXAMPLE.

In the plane triangle’ ABC

given the side a=2o,b=30, and

ez 4o required angle A.

RULE I.—SEOONn METHOD

,wrrnour HAVEESINES.

Three sides of aplane triangle being given, tofind an angle.

Put down the two sides containing the required angle,

and take the difference, under which put the third sidetake the sum and difference, and also the half sum and halfdifference.

To the arithmetical complements of the logarithms of thefirst two terms in this form, add the logarithms of the last

two, divide by 2, and look out the result as a log. sine the

angle corresponding to which will be halfthe required angle .1'

The sides of a triando are generally distinguished by the same

letters, but small, as the angles Opposite to them, as in the above

figure. The diagrams in this and the following problems are not

drawn to scale the method of constructing triangles by scale and

compass, although not giving such dorrect results as the logarithmic

method used in this book, ought to be known by the student. He

will find ample directions for solving triangles instrumentally in any0 work on practical geometry and mensuration .

1 This rule is derived from formula 0 (Part viz .

bc Sin.’ &(a— brw o) .

8022879.

8097940

0098970

A=2S° 57'15


EXAMPLES.

In the triangleABC, given a=5 l2, b c= 430

required C .

5 12 72 90730

627 72 02732

1 15 2 435366

430 2 197281

5 45 2)l9°126109

315 log. sin QC 9063054

2720 2 1°

26'45

”

1 570 2

and 5 3 30

In the triangle ABC, given a= °025 , b=§and c= °1 15 ; required B .

025 ar. co. log. 1 1002060

°1 15 ar. co . log. 10039302

090 log.

° 1075 1031408

°125 log. 0175 22 43038

2 15 2)l9°815808

035 log. Sin .-}B 9007904

°1075 §B=53°

59'

30”

0175 2

and B=107 59 0

In the above examples the angles have been taken out Of

the tables by inspection , namely, to the nearest If

greater accuracy be required, the student is referred to therules for proportioning for seconds which accompany histables of logarithms.

EXAMPLES.

Required the angles of any plane triangle ABC whosethree Sides, a, b, c, are given , namely

173. a= 798 Ans. A= 89° 45'

37

b=460 B : 35 1 2 7

c=654 C= 5 5 2 16

48 PLANE TRIGONOMETRY .

If the term marked be amiddle term : add the logarithms

of the two extreme terms, and subtract the logarithm of the

middle term not marked.

If the term marked be an w treme term : add the logar

ithms of the two middle terms, and subtract the logarithm

of the extreme term.

’

The result will be the logarithm of the required term.

See examples to Art. p. 16.

Note When two angles ofaplane triangle are known ,

the third can be found by adding together the twoknownangles and subtracting the result from 180

°the remainder

will be the third angle : hence, if two angles and the adjacent

side be given, the third angle is also known ,and theilce by

the above rule the other two Sides can be found.This rule depends on the property of triangles, that

the Sides are to one another in the same proportion as the

since of an angle Opposite to them and it will be in general

the easiest way ofworking, to write down such proportion ,and find the unknown term as in Art.

EXAMPLES.

In a plane triangle ABC, given a=50}, b=25 , and A68° 48

'to find the other parts.

Draw the triangle ABC, making the given parts as nearthe truth as possible without using instruments : mark the

given parts (A,a,b) of the triangle with a stroke of the pen ,

as in the figure then ,Since the

rule requires that the fourparts con

cerned Should be two sides and the

two angles Opposite to them,the

part to be found must be B ; markthiswith adifferent character(these

marks will render it more easy to apply the rule) then write

a am. A

PLANE .rmoONOME'rEr. 49

down a proportion as directed by the rule, and proceed to

find B.

The angles A and B being known, C is easily found

Note [1] and thence,by a proportion , the Side c.

To find B .

log. b 1097940

log. sin . A 90695 67

1 1067507

log. a

log. Sin. B 9066371

B=27° 38'1 5

To find the Side 8.

Draw the triangle ABC again ,

but mark the given partsA, a, and

C with a stroke of the pen, and the

required part c with a difierent

mark, as in the figure ; then makea proportion of the four parts marked thus

e za sm. C : sin . A

log. Sin. C 9097253

1 1098389

log. Sin. A 9069567

1 °728822

c=53°56

EXAMPLES.

Find the other parts ofthe triangle ABC, having given

a=214 Ans. B= 36° 6'30

b= 191 C=102 34 1 5

A=4I° 19' 15 c=316‘3

E

To find C .

27° 38'1 5

68 48 0

96 26 1 5

180

O=S3O 33'45


a= l7°25

c= 10ii orA=47° 0

'

30”

a= 96

b=§ A 28° 13'

B=45° 10' c= l '013

A=2O 10 C= 1 1 1° 35'

15 b= °5 41

a 4 b= -8743

a= °02 B=9O°

A= 1 1° 32' 15 c= °o9797

27 45

A=22 20 0 C= 108° 12'

27 30 a= 1

c=2§ b= 2

The Ambiguous Case.

When two sides and the angle opposite the less side

are given, each of the quantities sought will have two distinctvalues. For suppose that

,in the triangle ABC, the sidesBC

and AC and the angle A, opposite to the less side BC,are

given then ,if BC be not perpendicular toAB

, from C as a

centre, with radius CB, describe an arc cutting AB in another

point B and join CB,it is manifest that there will be two

27° 7'15

B= 105 5 2 15

b=22 ‘69

b=73'98

B=49° 0'

19

43

a=279 °7

b=243‘0

C=48° 0'

c= 2070

b= 265 '76

C=88° 2'

40

PLANE mIGONOMErEr. 5 1

triangles,ACB andACBI, having the sidesAC, CB, and angle

A,in one triangle, and the sides AC, CB,, and angle A, in

the other, of the same magnitude, while the remainingparts are different . Hence

it is evident that when such

parts of a plane triangleare given , the results willadmit of two different values. This may also be shown by considering that the angleABC orABlC is found by the formula a b Sin . A : sin. B ,

or a b Sin . A zain.ABIC therefore the Sines Of the angles

B and ABlC have the same numerical value (each being

b Sin . A

a

other. This also appears from the above figure for since

CB=CB,,the angles B and GBlB are equal ; thereforeABIC,

the supplement of OBIB,is also the supplement of B . To

find these two results we must, therefore, proceed asfollows

Having found theacuteangleB by the rule

, subtract itfrom 180°fortheangleAB,Cin the othertriangle there

maining ang lesACB,ACBl,A

and SidesAB,AB

”may then be found in the usual manner.

they are consequently the supp lements of each

EXAMPLE.

In the triangle ABC, or AB lC , given a=232, b=345 ,

andA= 37° 20’ to find the otherparts.

Calcu lation of angle B . There are two solutions.

In the first,

2 537819 B=64° 24’

82796 in the second,12020615 B

l= 1 15° 36

’

log. a 2065488

log. sin . B 905 5 127

First solution, B=64°24

’ Second solution, Bl= l l5°36

'

5 2 PLANE rmclONOME'rEr .

Calculation of angle C .

C 180° A— B

A 37° 20

'

B 64 24

101 44

180

C : 78 16

Calculation of side 8.

c za sin . C : Sin. A

2065488

9090829 log. sin. C

12056317

log. Sin . A

20735 21 0

c 3740

EXAMPLES.

Find the other parts of the triangle AB C , having given

19 1 . a 1780 Ans. A=54° or 125° 58'

b= 145 C =84 48, or 12 5 2

B=41° 10' c=2 19 °32 , or 4905

192 . a 25 978 4 B =80° 39

°

45 or 99° 20°

15

b=3084-33 C 43 7 30, or 24 27

A= 56° 12° 45 c= 2 136°7,or 1293-7

Rm III.

Two sides and the included angle being given, tofind theremaining angles.

Subtract the given angle from and thus find the

supplement divide the supplement by 2 . Put down the

two given sides,and take their sum and difference.

To the log. tangent of half the supplement of the given

angle, add the log. of the difference of the given sides and

Calculation of angle C.

C 180°— A— B

l

A 37 20’

B11 15 36

1 5 2 5 6

180

C : 27 4

Calculation of side c.

c : a Sin. C sin . A

2065 488

9058037

12023525

9 7 82796

22 40729

8 17407

PLANE TRIGONOMETRY. 5 3

from the result subtract the log. of the sum of the given sides;the remainder will be the log. tangent of half the differenceof the required angles which take from the tables.

To half the supplement of the given angle, add half thedifference of required angles just found, and the sum will be

the greater of the two angles required. To find the less of

the two rcquired angles, subtract the half difference.

In this case it is easier to work by the formula rather

than by the above rule deduced from it. Thus, if a, b, and

C be the given quantities, and the remaining anglesA and B

are required : by formula G (Part II. ) we have

a+ b

a— b — B)’

or expressed as a proportion

a+ h a— b tan. 1 (A+B) : tan .4} (A— B).

The first three terms in this proportion are known ; forhalf

the sum of the required angles, or 4 (180°— C)

(the half supplement of given angle) hence the fourth termis easily found, which gives us half the difference of the re

quired angles. The sum of the half sum and half difference

of the required angles will be the angle opposite the greater

of the two given sides and their diference will be the less

angle.

EXAMPLES.

Given a= 798,b=460,

andC= 5 5 ° 2'

required

the anglesA and B .

a+ b a — b : : tan. 5 (A+B) : tan . 4 (A— B)798 180° 0

'0"

460 5 5 2 15

1 258 2)l24 5 7 45

338 62 28 5 2

54 PLANE TEISONOMETEY.

log. tan. 4 (A+B) 102 83138 §A 16'

45

log. (a— b) 2028917 and 4}A+§ B=62 28 52

1201205 5 A=S9 45 37

log. (a+b) 3099681 and B=35 12 7

log. tan . 4 (A— B)

Find the other two angles of the triangle ABC, havinggiven

a=399 Ans. A=89° 45' 37

b=230 B=35 12 7

C=5 5° 2'15

b=64 B= 5 2 54 30

c=70 C=60 45 0

=66° 20’30

”

c=5 l2 8 1 1

b=627 58 1 1

C=42° 53'38

RULE IV.- F1EST METHOD

,USING HAVEESlNES.

Given two sides and the included angle, tofind the third side.

Add together 10002060, log. haversine of given angle,

and logarithm of each given side : take half the sum,from

which subtract log. of difference of the given sides. Look in

the tables for the remainder as a log. tangent, and take out

the corresponding log. Sine,which subtract from the half

sum just used. The remainderwill be the log. of the required

side.

Note. When the half sum is first found, write it down

If the student have no table of haversines, he may work the exples in this rule by finding, in the first place, the two other angles

by Rule and then the required side by Rule II. or he may pro

ceed as follows : Instead of using log. haversine, as directed in the

rule, take out twice the log. sine of half the given angle, and reject

the 10in the index of the constant log., then proceed as in the aboverule ; orhemay find the Side required by the nextmethod.

56 PLANE TEIGONOMETEY.

out the log. sine corresponding thereto, which subtract from

the log. under (2) The result will be log. of required Side.

*

In the plane triangle ABC, given a= 178,b= 145 , and

=84° 48'

required the third side c.

log. (a+ b) 2009203

log. (a— b) 10185 14

difference 0090689

log. tan .QC 9060530

log. tan . arc 1005 12 19

Find the third Side in following examples

196. a=798

b==460

C=5 5 ° 2'16 c=654

197. c=48

a=96

B=49° 0’

b=73‘98

198. a= 5 12

b=627

C= 42° 5 3'38 =430

RIGHT-ANGLED TRIANGLES.

Right-angled triangles may be solved by the rulesalready

given ; but the easiest way is to make use of the trigonome

trical ratios or definitions for the Sine,tangent, &c. given in

Art. p. 3, as in the following rule.

This rule is deduced from a formula investigated at p. 91 (Part

IL) , namely,

Sin . 0

[2]log. (a+b) 2009203

902885 5

sum 12038058

log. sin . arc 9097299

log. c 204075 9

side c= 219 °2

PLANE TRIGONOMETRY. 5 7

RULE V.

Right-angled Plane Triangles.

Select that expression or ratio in p. 3 which contains the

two known terms and also the term required, and when pos

Sible let the numerator of the ratio or fraction selected be the

part required. Reduce it to tabular logarithms, by Art.p . 25

,and then find the value of the unknown term in the

usual manner. See Ex. 1 17, p. 27.

EXAMPLES.

1 . In the right-angled plane triangleABC, given B=90°

,

a=42 , and A=5O°

required the otherparts.

Mark the given parts, namely A and a, with

a stroke of the pen ; and if c is the part about to

be found,mark itwith adifferent character, as in

the figure. Then

To find c.

By p. 3, ratio (d),2 cot. A,or c=a cot. A A

log. c=log: e + log. cot. A— 10

Calculation of c.

1023249

902 1247

1044496

c=35 °03

To find b.

Draw the triangle again ; mark the given

partsA and a with a stroke,and b

,the part re

quired, with a different mark as before then by

ratio (b), p. 3,gcosec. A,or b=a cosec. A

log. b=log. e + log. cosec. A— 10


log. a 1023249

log. cosec. A

log. b l °737938

b= 54 ‘7

To find C C=90° A=39O 50'

2 . Given a: 02, and B=9O° : required the other

parts. Make a triangle ABC,right-angled at B (see last fig. )

mark the given Sides a and c with a stroke of the pen, mark

also the part required with a different character.

Let the part required be A.

To find A.

By p . 3,ratio tan. A Z

log. tan . A— 10=log. a— log. c

or log. tan. A= 10+ log. a— log. c

2001030

10

8-301030

log. c 1000000

log. tan. A

A= 1 1° 18' 30

[2 ] To find Side b.

By p. 3, ratio 3sec. A

b=c sec. A

log. b=log. c+ log. sec. A 10

log. c 1000000

18’

30"

log. see. A

90085 14

10

10085 14

b= °102

find C C=90°— A=7S° 41'

30

PLANE Tmeorzouamr. 59

When two sides of a right-angled triangle are given, the

third may be found by the well-known property of rightangled plane triangles, namely, the square of the side opposite the right angle is equal to the sum of the squares of thesides containing the right angle

’

(Euc. i. thus, in a tri

angle ABC,right-angled at C, a2+ and if a= °02

,and

b: 1, we have

or nearly.

If one of the given sides is the hypothenuse, this method offinding the third side may be simplified : thus, in the aboveexample

, given c= ° 102,and a= °o2 to find b.

Since a2+b2=c2

b2=02

(c+a) (c x 082 : 010004

b= °1 nearly.

Find the other parts of the

A= 5 2O 38'0

b=45

B=9O°

A=49° 14'

c=33l

B=9O°

A= 5GO 29'

B=9O°

A=4O 44'

a=694 '73

B=90°

b= °2

=40°

B=9O°

c=04

C=40°

B=90°


206. a= 1777°5 Ans. A=56° 29'15

c= 1 177 30 45

B=90° b=2132 ’ 1

RULE VI.

Two sides and the included angle of a plane triangle being

given, tofind the area.

Add together the logarithms of the two given sides, and

log. sine of the given angle the sum,rejecting 10 in the

index, will be the logarithm of twice the required area.

*

EXAMPLE.

Given a=798, b=460, and C=5 5° 2

’

15 required the

log. sin. C

log. 2 area

2 area

and area

207. Given a=245 yards, b=760 yards, and O=GO°

required the area. Ans. 80627 square yards.

RULE VII.

Three sides of ap lane triangle being given, tofind the area.

From half the sum of the three sides, subtract each sideseparately. Add together the log. of the half sum and the

logarithms of the three remainders. Half the result will bethe logarithm of the

By formula I (Part II. )2 area=basin. A.

1 By formulaK (Part II.)

where s: }

2002003

2062758

9013563

5 478324

300832

150416

PLANE rmsorzonE'rar. 61

EXAMPLE .

208. Given a=798, b=460, and c=65 4 : required the

area.

area= 150418

209 . In the ABCD,AB=90

yards, BC= 100yards, CD= 1 10yards, DA_

120yards, the diagonal BD= 178'8 yards re

quired the area of the trapezium.

Ans. square yards.

2080458

21 98657

2095482

2 -480007

2)lO’35 4604

5 -177302

SPHERICAL TRIGONOMETRY .

PART I.

CHAPTER III.

RULES IN BPHEBICAL TRIGONOMETRY.

RULE VIII.— Fmsr METHOD,USING HALF HAVERSINES.

Three sides of a spherical"triangle being given, to find

an angle.


and take the difi’

erence,underwhich put the third side ; take

the sum and difference.

Add together the log. cosecants of the two first terms in

this form (rejecting the tens from the index), and the halflog. haversines of the two last termsfi‘ The result will be

the log. haversine of the required angle. 1

RULE VIII. — SE00NDMETHOD, WITHOUT HAVERSINES.

Three sides of a spherical triangle being given, tofind an

angle.


Aspherical triangle is that part of the surface ofasphere whichis bounded by arcs of three great circles, that is, three circles whose

planes pass through the centre of the sphere. The three arcs are the

sides of the triangles ; and any one of its angles is the same as the in

clination of the planes of the sides containing .the angle. For other

definitions in SphericalT rigonometry, see Trigonometry, Part II.

1 If the student have no table of half log. haversines, he may,after a little practice, take out of the table of haversines their halvesby insp ection or, the log. haversines may be written down on the

paper, and then divided by 2 . See example.

I By formula 0 (Part II.)

hav. A=cosse. b cosec. o J hav. hav. (a— b~c)

64 SPHERICAL TRIGONOMETRY.

0 0 0 0 0 0 0 0 0 0 0 0

Find the three angles of the spherical triangle ABC,

having given

RULE IX.— FIBST METHOD

,USING HAVERSINES.

Two sides and the included angle being given, tofind the

Add together 6301030, the log. sines of the given sides,d log. haversine of given angle reject 30from the index

of the sum,and take out the natural number of the resulting

logarithm. To this natural number add the natural versine

0000066

0023468

9078000

9098018

2)l 9°8995 5 2

log. sin . i A 9 -949776

§ A=62°58

’15

2

and A=l25 56 30

SPHEEIOAL TEIOONOIIETEY . 65

of the difference of the given sides the sum will be the na

tural versme of the required side,which find from the tables.

*

RULE IX .— SEOONDMETHOD, WITHOUT HAVEEsINEs.

Two sides and the included angle being given ,tofind the

remaining side.

Add together the log. sines of the given sides,

and twice the log. sine of half the given angle, rejecting 40from the index of the sum ; and take out the natural number

of the resulting logarithm. To this natural number add thenatural versine of the difference of the given sides ; the sum

will be the natural versine of the required side, which find

from the tableaxRULE IX .

— THIED METHOD (REQUIRING ONLY THE COMMON

TABLE OF SINEs,

Given two sides and the included angle, tofind the third side.

Take the difference between the two given sides : divide

it by 2, and thus get half the difl'

erence of given sides find

also half the given angle.

Add together the log. sines of the two given sides,and

divide the result by 2 , to which add the log. sine of half the

given angle. Call this sum log. M . From log. M subtract

log. sine of half the difference of given sides, and look in the

tables for this result as a log. tangent ; and take out the corre

sponding log. sine,which subtract from log. M . The remainder

will be the log. sine of half the required sidai

By formula, p. 94, Part IL,we have

vers. a=vers. sin . b sin . c hav. A.

1 Also, at same page,

vers. a=vers. sin . b sin . c sin 3

1 By formula, p . 95 , Part‘

II.,

g_ sin . 5 AJ ein . b sin . c

2 sin 0

Bin . 5 (5M ?)

66 sPHEaIOAL TRIGONOMETRY.

Note. When log. M is first found, write it down again a

little to the right hand, for the convenience of placing under

it the log. sine to be subtracted .

EXAMPLE.

In the spherical triangle ABC, given c= 1 19° 42'20

b=108" 4' and A=75 ° 31 ’required the remain ing

side a.

c= 1 19° 42’20

b= 108 4 18

c— b 1 1 38 2

Eat. ver. (c— b)

nat. ver. a

By Second Method.

c= 1 19° 42'20

'

b= 108 4 18

c — b 1 1 38 2

5A 37 45 45

nat. ver. (c— b)

nat. ver. a

log.

619349

205 44

639893

a=68° 5 3'

36

const. log.

log. sin. 0 9038818

sin. b 99 78031

sin . 5 A.

sin . 5 A.

log. 5 -79 1935

619349

205 44

639893

=68° 53'

36”

sPHEEIOAL TBIGONOMETBY. 67

By Third M ethod .

b= 108°

log. sin . b 9078031

c=1 19 42 20 log. sin. c 9038818

38 2 2)l9-9 16849

49 1 9058424

log. sin. 4A

§A= 37 45 45 log. M .

log. sin . 9005805

log. tan. 107 39647. log. sin . 9092916

log. sin . 5a 9 75 25 36

§a=34°

26’

45

2

side a=68 53 30

EXAMPLES.

Find the third side of the spherical triangle ABC, havinggiven

2 14. A 0”b==76 42 0

oz 89 10 30 a= 96°9'5 7

"

2 15 . A= 50 0 0

b: 70 45 10

c='

62 10 15 46 19 32

2 16. a= l00 8 42

b: 98 10 5

C : 88 24 24 87 0 50

217. b==1 18 2 14

c= 120 18 33

A 27 22 34 a= 23 57 9

218. a: 87 10 15

b: 62 36 45

= 102 58 30 c= 100 9 38

a= 69 19 10

b 78 59 14

C= 1 10 48 42 = 104 59 5 7

68 sPHEEIOAL TmGONOHETar.

RULE X.

Of two sides and the opposite angles, any three being

given, to find thefourth.

Write dow naproportion having for the two first terms

the sines of the two sides concerned, and for the third term

the sine of the angle Opposite to the first side pu t down , and

for the fourth term the sine of the angle Opposite to the other

side put down in the proportion . Mark the term required ,and proceed as in the corresponding rule for plane triangles

(p.

EXAMPLES.

In the spherical triangle ABC, given a=70° 10’30

b=80° and B=33° 15' required the angle A.

b= 1 1 9° 42'

a= 108° 4'18 and A=99° 21'30

sin . a : sin . b : : sin . A zsin. B

log. sin . a 9073466

sin . B 9 739013

19 71 2479

sin . b 9093462

sin . A 9 °7l9017

A= 31° 34' 30"

In example a being less than b,A is less than B,and

therefore A is less thanIn example b being greater than a

, B is greater thanA, and therefore B is greater than 90° hence the angle

taken out of the tables must be subtracted from 180°to getB .

Unless by some limitation of this kind it should be determined that the required angle is less or greater than

By formulaN (Part II.)sin . a sin . A

sin . 5 sin. B

sin . a sin . b : sin .A : sin. B

log. sin. b 9038818

9094181

19032999

sin . a 9078031

sin . B 905 4968

B= 180° — 64° 2 1’

30

— 1 15°38

'

30"

sPHEaIOAL TRIGONOMETRY. 69

this rule is considered ambiguous. See corresponding rule

for plane triangles (p.

RULE XI.

Two sides and the included angle being given, to find theother two angles.

Take the sum and difference of the two given sides, andthe halfsumand halfdifference, and also half the given angle.

Under heads [1 ] and put down the following log

Under [1] and log. cot. of half the given angle.

Under log. cosec of half the sum of the given sidesUnder log. sec. (rejecting 10in each index) .Under log. sin .

Under log. cos.

Add together the logarithms under [1] and and (rejecting10 from each index) the results will be the log. tangents ofhalf the difference and half the sum of the required angles

espectively, which take from the tables (a) .The sum Of the two arcs thus found will be the angle op

posite to the greater of the given sides and their differencewill be the angle opposite to the less of the given sides.

*

(a). Note. If half the sum of the given sides be greaterthan half the sum of the required angles will also be

greater than 90° in this case, the are found as above under

[2] must be subtracted from 180° to get half the sum of the

required angles.

ofhalfthe difference ofgiven sides.

EXAMPLE .

220. In the spherical triangle ABC, given air- 124°

By formulaT (Part II.)

tan .

tan . (A cot. aC

(a— b)

70 SPHERICAL TEIOONOM'ETEY.

b=89° 0'

C= 1 12° 1’

required the other twoangles,A and B .

a= 124°10’ 0 C= 1 12° 1

’30

b== 89 0 15 QC== 5 6 0 45

a+ b=.

213 10 15 35 7

a— b= 35 9 45 § (a— b)= 17 34 5 2

[1 ]cot. QC 9 828783 cot. QC 9028783

b) 001845 5 sec

- b) 4k(a— b) 9079225

tan. &(A— B) 103 52488

&(A— B) 1 1° 5 9'

45 arc 66° 3’

180

1 13 5 7 0

and } (A— B) 1 1 5 9 45

5 6 45

B= 101 57 15

221 . Given b= 89° 0’ o= 108° 56’45

”

to find the angles B and C .

Ans. B= 101°5 7

’

15 C= 1 12° 2’

Examples to this and the following rule are easily formed

out of those already given asEX. 21 1 , &c.

RULE XII.

Given two angles and the included side,tofind the other

Take the sum and difference of the two given angles, and

the half sum and half difference, and also half the given side.

Under heads [1] and [2] put down the following logar

Under [1] and log. tan. of half the given side.

The are 66°3

’

is subtracted from because‘

the sum of the

sides a and b is greater than (See note, p.

72 sPHEaIOAL TEIGONOMETEY.

RULE XIII.

RIGHT-ANGLED SPHERICAL TRIANGLE .

In a right-angled spherical triangle ABC, A being theright angle, the two sides adjacent to the rightangle 5 and c, and the complemen ts

"of the three

other parts, namely, co. a, co. B, and 00. C ,

are

called thefive circularparts. By the following

rules, called, from the name of their author,Napier

’8 Rules, two of these parts being given,

any one of the other three may be found.

In applying the rules, we must select one of

the three parts concerned, such that the other

two may either be both adiacent to it, or both separate from

it. The part so selected is called the middlepart. RuleA

will apply to the former case, Rule B to the latter.

RULE A.

The sine of the middle part is equal to the product of the

tangents of the two parts adjacent to it.

The complement of an angle iswhat itwants of thus, in the

P right - angled triangle PNC

( 1st the angle P, or

90°— A, is the complement of

the angleA. It may easily be

proved that the sin., tan .

, sec.,

cos., cot., and cosec. of an

angle are the cos., cot., cosec.,

sin., tan., and sec. of its com

plement respectively. Thus,

let the two trianglesCFH be equal to each other in every respect, thenP is the complement ofA, or co. A.

Now [Art. p. 3] sin. A cos. (90°

co. A

a

lig cot. (90°

co. A, 650.

Hence, in the above rules, sin . A is substituted for cos. 00. A ; cos. A

for sin. co. A; cot. A for tan. 00. A, &c. (See Part II.)

sPHEaIOAL TEIOONom ar. 73

RULE B.

The sine ofthe middle part is equal to the product of the

cosines of the two parts opposite to, or separated from it.’

Having written down the equation according to the case,make a dash under the part required, and determine its magnitude by applying the proper signs or to each term

(Rule XVI. p.

In a right-angled spherical triangle ABC, given 5

74°19

'

o=38° and A=90° : required.

the otherthree parts.

Make a triangle for each of the three required parts, andmark the two given parts and the unknown part as directed

(p. and then consider which I. 0

of the three is the middle part,and whether the other two marked parts are adjacent

,or opposite

parts.B c A o A

Thus in fig. 1

,to find B, the right angle A not being con

sidered as a part, 0 will be the middle part, and complement

B and bare the adjacent parts.

By Rule A, sin . c=tan. co. B . tan. b

or sin . c=cot. B tan . b

(since tan. co. B=cot. B by note, p. 72) and determiningthe sign of B by Rule XVI. p . 30, we have

sin. c cot. B tan . b

log. sin. c— 10=log. cot. B — 10+ log. tan. b— lo

whence log. cot. B=log. sin. c lo— log. tan . b

Forverification of these rules, see p. 72, Part II.

74 SPHEEIOAL TRIGONOMETRY.

Calculation.

b=74° 19 ’30 log. sin . 0+ 10

o=38 5 6 tan . b

cot. B

B=8O°

In fig. 2, to find C . Marking the figure in the usual

manner, we see that b is the middle part, and c and C the

adjacent parts.

By Rule A, Calculation.

sin . b=tan . c tan. co. C log. sin . b+ 10 19083540

9007336

or sin . b=tan . c cot. C 10076204

C=40°

In fig. 3, to find a. In this case a is the middle part,and b and c are the parts separated from it.

By Rule B,

sin . co. a=cos. b cos. c

or cos. a=cos. b cos. c log. cos.

To find a (fig.To find 0 (fig.

cos. a=cot. B cot. 0

log. cot. B 9 321222

cot. C 9047508

008. a 92 68730

79°18

'0

”

180

a=lo0 42 0

Calculation.

log. cos. b 94 31654

cos. 6 9 8909 1 1

a (rejecting 10) 90225 65a=77

° 5 2"

Given A= B

— 78°10

’

and C131

°

32'

to find

the other parts.

To find b (fig.

cos. B=cos. b sin. 0 cos. C=sin . B cos. c

cos. B+ 10. 1901 1893 cos. C+ 10 1902 1657

sin . 0 9074148 sin . B 9090671

5 9 437745 cos. 0 9 830986

b=74° 5' 45 47° 20'30

"

180

c=132 39 30

sPEERIOAL TEIOONOMETar. 75

In a right—angled spherical triangle ABC, find the otherparts, having given :

223. B : 72°

l9’0

”

5 : 50 50 0

A: 90 0 O

b: 60 10 0

0 0

A_ 90 o 0

225 . B= 100 0 0

C : 87 10 0

A: 90 0 0

226. oz: 46 18 23

B : 34 27 30

A: 90 0 0

227. c= 1 18 2 1

A 23 40 12

B : 90 0 0

228. a= 100 42 0

B : 78 10 0

A= 90 0 0

229 . e 53 14 20

A= 91 25 58

B : 90 0 0

RULE

QUADEANTAL sPHEBIOAL TRIANGLES.

A spherical triangle having one side 90° is called aquad

rantal triangle.

The five circular parts in a quadrantaltriangle are the two angles adjacent to

the quadrant, and the complements of the

otherthree thus, ifABC be a quadrantaltriangle, the side a being the five

circular partsare the angles adjacent to the

quadrant B and C, and the complements B

of the angle A and of the sides b and c.

a= 54°28

'0

”

23 2 15

C 28 45 0

B : 60 32 45

C : 98 41 45

a= 94 5 7 15

a= 90 30 0

b= 100 0 45

c= 87 7 1 5

b: 26 23 15

a 5 1 46 15

C= 66 5 9 30

b= 1 16 18 0

C=100 59 30

a 2 1 5 45

C= 131 32 45

c= l32 39 30

b 74 5 45

b= 91 4 15

C : 53 15 0

a: 91 47 15

76 SPHERICAL TEIOONOMETEY.

Any two of these parts being given, a third may be found

by the preceding rules for right-angled triangles (p. 7

The magnitude (whether greater or less than -

oi the

part required is determined by Rule XVI. p. 30, observing,that when two angles or two sides come together o nthe same

side of the equation, the sign must be placed before them,

and the three signs thus placed on one side of the equation

must be made to produce the same result (positive or negative) as the sign of the other side.

“

In the quadrantal triangle

ABC, given B = BQ° C

48°

and a= 90 : find the

A other parts.

To find A (fig. To find b (fig.

cos. A — cos. B cos. 0 sin . C=tan. B cot. 5

log. cos. B 9032444 sin . 0+ 10 19876678

cos. 0 9018392 tan . B lo°

76l l28

008. A cot. b

83°32

'45

”b=82° 34’

180

A=96 27 15

In a quadrantal spherical triangle ABC,find the other

parts, having given :

230. a: 90° 0’

0

A= 100 0 0

c 50 10 0

231 . B 45 0 0

c 72 0 O

a: 90 0 0

For verification of rules for quadrantal triangles, see p. 74,Part II.

To find 0 (fig.

sin. B=tan. C cot. 0

sin. B+ 10 190935 72

tan. 0 10058287

cot. o . 9035285

c=49° 15 '5 1"

SPHERICAL TRIGONOMETRY. 77

78 EXAMINATION- PAPERS .

EXAMINATION PAPERS IN PRECEDING RULES.

I .

237. Required the product of 18 x 48 x 6 2 x 4

Ans . 21427

238. Divide 236 by 161239 . Required the 3d power of 12 5 195 3 125

240. 4th powerof°076543 000034326

241 . 6th root of 1 1078 4 72 146

242. 7th root of 098674 7 183146

243. value of 3-247

244. value of a: in the following proportion24 79 zaz Ans. x=5 7°604

245 . Reduce the following expression to lOgarithms

9

3 Ans. log. x=2 log. u+ log. b— log. c— log. d

246. Find the value of a: in the following equation

Ans. m=9 ’ 5868

247. The first term of a geometrical series is 2, its com

mon ratio 3, and number of terms 8 find the sum.

Ans. S=6560

248. If 4001. be placed out at compound interest for nine

years, at 4l. per cent. perannum,required its amount.

Ans. 5691. 6s. 8d.

11 .

249 . In the plane triangleABC, given a= 10, b= l5 , andc=20; required the angleA. Ans. A= 2S° 57

’

15”

250. In the plane triangle ABC,given a=25 ° 125 ,

and A=GS° to find B . Ans. B=27° 38'

15“

25 1 . Given a=399, b=230, and C= 5 5°2'

requiredthe angle A. Ans. A=S9° 45

'

37’

252 . Given a=40,b=60

, and C= 100°

required 6.

Ans. c= 77°68

253. In the right-angled triangle ABC, given B=90°,a=210

, and A=5O°10

’required the other parts.

Ans . C=39° c= l75 ° 1,b=273‘5

80 EXAMINATION ru n s.

267. In the quadrantal triangle ABC, given a=9o°,B= 74° 36

'and c= 50° required the other parts.

Ans. A= 100°, b= 78°14

'

C= 49° 8’15

268. Given a= 90°, = 96° 17'

b=82° required

the other parts. Ans. B=80° C= 50° c= 50° 27

’

V.

269 . Required the product of 2 -4 x 0035 x 108 x -1 .

Ans. 0009072

270. Divide 95 by '36. Ans. 26 389

84 x 0769 x 00683Fmd th"“ 13° °f

5 98 x 0000146 x 0039

Ans. 1295 71

271 . Find the 3d power of 0321 and the 72d power of

06797. Ans. 0125 and 096

272 . Required the 200th root of0063241 . Ans. 075

273. Required the value of “lll0958825 . Ans. 00563

274. Required the value of the following expression

W 72

Ans. 2 9904

275 . In the plane triangle ABC, given b= °2705 ,

and c= ‘3375 ; required the angle C . Ans. C= 1 1 1° 47'45

”

276. In the plane triangleABC ,given a= 1 16, b= 1 72'5

and A= 37° to find B . Ans. B= 64° or 1 15 ° 36

277. In the plane triangleABC, given a= 5 l2,b= °627

,

and C= 42° 53'

required the angles A and B.

Ans. A= 5 4° 8’ = 82° 58

’

1 1

278. Given b= °2,c= °25

, and A= 22°

required the

third side a. Ans. c= 'l

279 . In the right-angled plane triangle ABC, given a=

177i c= 1 17°7, and B=90°

; required the side 5 .

Ans. b= 213°2l

280. In the trapeziumABCD (fig. p. given the sideAB=90 yards, BC= 100 yards, CD= 1 10 yards,DA= 120

yards, and the angle DAB= 1 16°

; required the area.

Ans. Area=9768°7 yards

CHAPTER IV.

APPLICATION OF THE RULES AND FORMULE OF PLANE AND

SPHERICAL TRIGONOMETRY TO THE SOLUTION OF PROBLEMS

IN ASTRONOMY, SURVEYING, AND NAVIGATION.

When a line joining any two points in space is

accessible throughout its whole extent, it may in general be

measured by the successive application of some line of a

known length but when it is inaccessible, or cannot be di

rectly measured, we may obtain its length by considering it

the side of a triangle, if we already know , orcan find byoh

servation, asufficient numberof parts of that triangle to enable

us to apply one or more of the preceding rules. It is thus

that the mensuration of inaccessible lin es of any length is

found by means of that of accessible lines and angles.

The angle DOC, contained by lines drawn from a point

0 to two remote objectsC and D

, may be mea

sured by placing a cir

cle in the plane passingthrough the two objects,and having its centre atthe angular point 0. The straight edge ofa ruler being thenplaced on the circle so as to pass through its centre, and, bymeans of sights placed over it, directed first to one Object andthen to the other, the arc AB of the circumference betweenthe two positions of the ruler can be found this is the measure of the angle 0.

The principal instruments formeasuring angles arethe theodolite and sextant. A theodolite is the most convenient instrument formeasuring horizontaland verticalanglesit is composed of two circles having their planes perpendicu

G

82 PROBLEMS IN SURVEYING.

lar to each other. When the instrument is used, one of the

circles is placed in ahorizontal plane, bymeans of levels ; on

this circle horizontal angles are measured ; on the other

measured vertical angles, whether of elevation or depression

(that is, whether the object is above or below the horizontal

line) . A Sextant is employed tomeasure angles contained in

any plane whatever. It is more suited for Observing angular

distances of heavenly bodies than the theodolite but the

latter is better adapted for Surveying than the former, since

it determines the horizontal angles at once ; but those Oh

served with the sextant must, when out of the plane of the

horizon, bereduced to that plane by calculation, to suit them

to the purposes of the survey.

We will not stop to give particular descriptions of the

by a careful study of the instruments themselves ; we shall

therefore suppose the manner ofadjusting andapplying them

to practice is known,“ and proceed to give acollection ofpro

blems in which these instruments have supplied for the most

part the necessary data.

SEU'I'ION I.

Problems in Surveying,&c.

The rules of Plane Trigonometry just given will enable us

to solve many useful and interesting problems in SurveyingandNavigation. A large collection of these will be found at

the end of the book ; they have been selected to serve as an

introduction to Navigation and Nautical Astronomy. We

will give, however, in this place a few that most frequently

occur in practice, with their solutions at length and as it is

often very necessary that a naval student should know how

to solve trigonometrical problems, not only by logarithms,

The adjustments and method of using the sextant, the instru

ment chiefly used by the seaman for measuring angles, are fully de

scribed in Navigation, Part I.

84 PEOELEMs IN SURVEYING .

distance from a ship at anchor at Spithead, I observed witha sextant the angle between the ship and the steeple of St.

Thomas’

s Church to be 72° I then walked 500yards in

a direct line towards the church, and again took the angle

between the ship and the steeple, and found it to be 82°

Required my distance from the ship at each observation.

Let C be the ship at anchor, A my first station, B the

second, and P the church. Then the angle CAP=72°

CBP==82° and the line AB=5OOyards, are given to find

AC and BC, my distances at each observation .

In the triangleABC, theangleABC= 180°

- CBP=97°

15'

angle C= 180°

To find BC .

By Rule, p. 47, BC zAB : : sin. A : sin. C .

=500 log. AB

A=72° 42' sin. A

C=10 3

To findAC.

AC :AB : sin. ABC zsin . C

2098970

9079895

12078865

sin. 0 92 41814

BC 3-43705 1

distance BC=2735 yards.

PaonLnns IN SURVEYING. 85

AB=5OO 2098970

ABC=97° 15' sin. 90965 14

3 12 695 4849 -241814

AC 34 5 3670

distance AC=2841 yards.

3. Being ordered to place a target at 500 yards from the

ship, and knowing that the height of the truck above thewater-line was 213 feet, it is required to find what angle thisheight will subtend on my sextant when I am at the requireddistance (before allowing for index correction of instrument).

Let BC represent the ship’

s mast,A the required place of

the target : then the angle BAC is the angle which must beread off on the sextant (supposing it to have no ind. cor.

In the right-angled triangleABC are given the sideBC2 13 feet, AB=5OO yards, or 1 500 feet, and B=90

°to cal

culate the angle A.

By Rule, p. 5 7, tan . A

BC : 213 log. tan A— 10=log. BC— log. AB

AB= 1500 or log. tan. A= 10+ log. BC — log. AB

log. BC 10

3°l76091

9 °l5 2289

=8° 5'

to keep at the distance of 500 yards from her,and knowing

that the height of hermast above the hammock nettingswas

86 PnonLnus IN sUEm ING.

198 feet, it is required to find what angle on my sextant willindicate the proper distance.

LetAandB be the two ships,Athe place of the observer,and CAB the angle subtended by the mast CB at A. Then

in the right-angled triangle CAB are given CB=198 feet,AB= 1500feet, and B=90

°to find the angle CAB.

By Rule, tan. ACB

log. tan. A— 10=log. CB— log. AB

log. tan . A=10+ log. CB— log.

log. CB+ 10 12096665

31 76091

9 1 20574

=7° 31' 15

5 . Wishing to determine the height of a lighthouse C on

the summit ofa did on the seashore, I observed the angle of

elevation CAD of its top above the level sand tobe 26°

88 PROBLEMs IN SURVEYING.

Let CD represent the mountain, A the ship’s place at

the first Observation, B the place at the second Observation .

Then, in the triangles CAB and OBB we have given the side

AB=3°5 miles, the angle CAB= 12°

and the angle

CBD=30° 13’

to find the height of the mountain CD,and

the distance DB from the station B.

TO find BC .

In the triangle ABC, BC AB sin. BAC sin . AOB.

AB=3'5 log. AB

BAC= 12° 25’

sin . BAC 9032478

ACB=CBD— BAC 9 8765 46

CBD=30° 13'

BAC= 12 25

ACB= 17 48 BC=2 °46 miles.

(2 To find CD.

In the right-angled triangle CBD,sin. GED

CD=BC sin . CBD

BC=2 '46 log. BC

CBD=30° 13'

sin . CEB

CD 009305 9

CD= 1 '24 miles.

(3 To find DB.

BDIn the right—angled triangle CBD, BC

cos. CBD

BD=BC cos. OBB

PEOBLEMs IN sUnva NG. 89

BC=2 '46 log. BC 009125 7

CBD=30° 13' cos. CBD 9036578

BD 0027835

BD=2 '13miles.

7. Wishing to know the distance between two ships at

anchorat C andD,I measured on the shore abase lineAB

735 feet, and with a sextant Observed the following angles.

D

AtA,one end ofthe base, CAD=63

°andDAB=35 °

at B,the other end of the base, DBC=80

°and CBA

28° Required CD, the distance between the two ships.

In triangleACB,find AC .

In triangleABD, find AD.

In triangleACD, find CD.

To find AC.

In triangle ACB are given AB=735 , CBA=28°

CAB=63° and therefore the remaining angleACB=53

°

CBA= 28° 20’

CAB : 98 40

127 0

180

AGB= 5 3

AB=735

:AB : : sin. CBA : sin. ACB

log. AB 2066287

9076328

12042615

9002349

2040266

AO=436°8

90 PEOBLEM8 IN SURVEYING.

To find AD.

In the triangleABD are given AB=735 ,DAB=35°

ABD=28° and therefore the re

maining angle ADB=36°

DAB : ADB

ABD= 108 36 log. AB 2066287

143 46 sin. 9076702

180 12042989

ADB 36 14 sin. ADB

AB=735 3071346

AD= 1178°5

(3 ) To find CD .

In the triangle ACD are given AC=436°8,AD= 1 178°5

,

and the included angle CAD= 63°

By Rule IV. second method, p. 5 5 .

1 178-5 63°30

'

AC 436-8 31 45

AD+AC 1615 -3

AD — AO 741 -7

[2 ]log. (AD 3008253 log. (AD 300825 3

(AD 28 70226 9 -72 1 162

0338027 12029415

ad i eu ) 9 79 1563 9004757

tan . arc. 10-1295 90 3024658

CD=1058‘5

As the two following problems are ofgreat use in Marin eSurveying, we will solve them by logarithms, and also by a

geometrical construction . In problem 98 of the volume of

Astronomical Problems, analytical solutions of the same

problems are also given .

8. Wishing to determine the position ofa sunken rock at

92 PEOBLEM8 IN SURVEYING.

Assume any point G to be the position of the boat, andlet A,

B,and C be the Objects. Describe roughly a circle

passing through the three pointsA, G,C . Join GA, GB,

and

GC . Then GA, GB,and G0are the distances required. Draw

Al), CD to the point of intersection D . Then ,by Geometry,

since the angles in the same segment of a circle are equal,

CAD=CGB=34° and ACD=AGB= 26°

FindAD,having given in the triangleADC the side

AO=7, the angle ACB=2G°

andADC= 180° 12'

+ 26°

Find angle BAC, having given the three sides of the

triangle ABC.

[3 ] Find angle ABD, having given AB ,AD

, and angle

BAD (=BAC — CAD) .Find GA, GB,

and angle BAG, having given in the

triangle ABG the side AB and the anglesAGB,and ABG.

Find GC, having given in the triangle AGC the

side AC, the angle AGC, and the angle CAG(=BAGBAG).

To find AD.

AC :AD : : sin. ADC :sin.DCA

AO=7 0045098 AB=5

ADC=1 19° 21’

9048766 BC=6

DCA= 26 27 104 93864 AO=7

9040338

0053526

AD=3°577

To find angleABD. Rule p. 5 2.

To find angle BAC

7 9 -154902

5 9001030

2 0-602060

6 0001030

8 905 9022

4 BAC=5 7°

4 CAD 34 12 0

2 BAD=22 5 5 15

PEOBLEMl IN SURVEYING . 93

AB=5

AD=3°577 10092995

8077 10046200

0033335

BAD: 22°

5 5’

15 9012865

180 39°17

’

15”

15 7 4 45 78 32 22

78 32 22 ADB= 1 17 49 37

ABD: 39 15 ‘7

[4 ] To find GA,GB

,and the angle BAG.

AB=5 AB GA sin . AGB sin . ABG

AGB : 26°27

' AB : GB sin. AGB sin. BAG

ABG : 39 15 0098970 0098970

65 42 9001201 905 971 1

180 10000171 10058681

BAG= 1 14 18 9048766 9048766

005 1405 1009915

GA=7°103 GB=10°23

[5 ] To find GC .

AC : GC sin . AGO : sin. CAG

0045098

AGC= 60°39

’902449 1

BAG= 1 14 18 10769589

BAC= 57 7 9040338

CAG= 57 1 1 002925 1

GC=6°75

9 . Sailing in a deep and unknown bay, I suddenly foundthe soundings decrease ; and suspecting I was on acoral reef,I hauled 011,having anchored a boat on it, from which the

followingangleswere taken between three remarkable objects,A, B,

and C, that appeared on the distant shores, namelybetween Aand B, a high-pointed rock to the right ofA, the

angle was 1 16°40

' between B and C, a bluff summit to the

94 PEOELEns IN sUEm ING.

right of B,the angle was 1 12

° I Obtained afterwards thedistances between the three pointsA,

B, and C, by mensuring on the shore convenient base lines, and taking angles, aspointed out in Example

,p. 89 . The distance from A to B

was miles,from B to C 75 miles, and from A to C

miles. It is required to find the position of the reef.

(1 By Construction .

Construct the triangleABC from the scale of equal parts,by taking AB=5

'75 , BC= 7'5, and

AC=8°25 . At the point C, in thestraightlineAC ,

makeACD 63°

the supplementoftheangle subtendedC by the other two points A and B .

Again , at the pointA, in the straightline AC, make CAD=67

° the

supplement of the angle subtendedby the other two points B and C

produce the linesADand CD tomeet

in the point D . About the triangle

ADC describe a circle then th8

place of the reefwill be somewhere in the circumference of

this circle. To find it, join BD and the point of intersec

tion G is the position of the reef required.For since the angles in the same 8egment of a circle are

equal (Euclid, b. therefore AGD=ACD=63° 20’

therefore the angle AGB= 1 16° Again, CGD=CAD=

67° 30’ therefore the angle BGC= 1 12

° And thesewere the angles Observed at the boat ; therefore G must bethe place of the boat, or position of reef.

Assume any point G as the position of the reef, and let

A, B,and C be the objects

s

on shore. Describe a circle passing through the three points A, G, and C. Join BG,

and

produce it to meet the circle in D . Join GA and GC .

Then GA, GB, and G0are the distances required. Join AD

PEOELEME IN NAVIGATION.97

Application of Plane Trigonometry to

THE OOMPAss.

It is proved in Navigation, Part II , that the course,distance, true difi

'

erence of latitude, and departure between

any two places on the surface of the earth, mayrepresented by the sides and angles of A

a right-angled plane triangle. Let us

suppose A and B to be the two places,and AC

‘

that part of the meridian pass

ing through A that is intercepted be

tween A and a straight line BC drawn

through B perpendicular to AC . Then

(see Navigation, art. 49) AC will re

present the true difi’

erence of latitude,AB the distance, -and BC the departurebetween A and B, and the angle CAB

the course HourAto B. It is manifest,

that if in the triangle BAC any two of the above quantitiesare given , the other two maybe found by the common rule forright—angled plane triangles. We will exemplify this bymeans of a few examples in sailing, and at the same time will

98 PEOBLEirs IN NAvIGATION.

show how to find the ship’s place by means of construction,

that is,by the use ofmathematical instruments the practice

thus obtained will be useful\

to the Naval student, and willform a proper introduction to the construction of charts and

the tracing the ship’

s track thereon .

But before we can construct the ship’

s track for differentcourses

, the several points of the compass must be thoroughly

known . An expeditious method offorming the compass,and

of learning it, is given in Navigation, Part I. p. 1 1 .

10. A ship from latitude 47° 30’ N. has sailed98 miles find by construction , and by calculation, the latitude she is in, and the departure she hasmade.

By Construction .

Let Arepresent the point the ship departed from,ADthe

meridian,andAp ,

drawn at rightangles to it, the parallel Of lati

tude of the ship. At the pointA, with the chord of de

scribe the quadrant mp , and clitOff or 33° 45

'

the course and through c draw

a line AB . From a scale of

equal parts take AB=98miles,mthe distance and thrOugh B

draw BD parallel to Ap ,meeting AD in D . Then B is the

place the ship has arrived at,AD is the difference of latitude,and BD is the departure. If AD and BD are measured by

the same scale Of equal parts,it will be found that the differ

ence of latitude AD is about 81 miles, and the departure BD

about 5 4miles. The figure may be more easily laid off by

means of aprotractor (see any work on PracticalGeometry).

By Trigonometry.

In the right-angled triangle ABD are given the course

DAB=33° and distance AB=9S miles to find the dif

ference of latitude AD and departure BD.

100 PROBLEMS IN NAvIGATION.

rants,WS, into eight equal parts to formascale ofpoints. Thismay be done by bisectingWS in a, and then Wa in b, and

Wb in c; the otherpoints, d efg, may then be readilyfilled in.

Mark 03 on the circumference the'

several courses,thus : take N1=N.N.W .

,or two points from the scale in

WS and or one point from the east.

Through A1 draw the straight line AB, and make

AB=90miles by a scale of equal parts and through B, par

allel to a line passing through A9 , draw BC=60miles. The

point C represents the place the ship has arrived at. Join

AC , and through C draw CD parallel to WE, meeting AN

produced in D. Then AD is the difference of latitude, and

DC the departure madegood during the day. Also the angle

CADand line AC represent the direct course and distance

fromA to C .

Ifwe measure AD by the scale of equal parts, we shall

find the difference oflatitudeADabout 71 milesto the north,and the departure DC about 24 miles to the east ofthe place

the ship left. The latitude arrived at is found thus

Lat. A 47°

30'N.

Diff. lat 1 1 1 N.

Lat. in 48 41 N. and dep. 24 E.

By Trigonometry.

To calculate the difference of latitudh and departure be

tween A and C, we must proceed as follows

Through B draw BF parallel toWE,meeting themeridian

produced in F. Then in the right-angled triangle ABE are

given the course BAF=2 points, and distanceAB : 90miles,

to calculate AF the diff. lat. and BF the departure. Again,through B draw BG parallel to the meridian NS and through

C draw CG parallel toWE, meeting BG in G . Then in thetriangle BGC are given the course GBC=7 points, andBC60miles, to calculate BG : FD the diff. lat. and G0 the de

parture. By performing the calculation , we find that AF:

83 2 miles to the north,and BG :

°

1 1 '7 to the south ; so thatthe diff. lat.=83'2 N.

— 1 1 -7 s.=71 -5 miles. Similarly may

mom ma IN NAVIGATION. 01

be found GC=58°8 to the east, and EB orDC=34 °4 to the

west so that the departure=58°8 E. W .=24°4 E.

Thismethod ofcomputing the diff. lat. and departure separately for every course is in practice avoided by making use

of a table called the Traverse Table, which contains the diff.lat. and departure, calculated for any given course and dis

tance, so that these quantities may be found by inspection .

The difll lat. and departure,when taken out of the table, are

arranged under proper heads, in the following form

N. 244 E .

N.

1 1 1 -5 N.

Lat. in 48 41 -5 N. and dep. 244 E.

A ship in latitude 32° 14’ N. has sailed during theday S.S.W . 45 miles, and W.b.N. 30 miles. Find latitudein, and departure. Ans. lat. in 31

° 38’ N. dep.

12 . A ship from latitude 50° 48'N. has sailed during theday on the following courses. Required the latitude in

, and

departure, and thence her direct course and distance from one

place to the other.

1 . SE . 40miles. 4. N

2 . NE. 28 5 . S.S.E

3. 5 2 6.

(l . ) By Construction.

Let A be the place sailed from, and NWSE the horizonof the ship. Draw the meridian NS, and parallel of latitudeWE.

Divide one of the quadrants into eight equal parts for a

102 PROBLEMS IN NAVIGATION.

scale of points, as in the last example, and by means of this

scale mark off on the circumference the several courses, viz.

$ 5

Nand

Through A draw AB=40by a scale of equal parts ;through B,and parallel toA,,

draw BC= 28miles; through

C, and parallel to A5 ,draw

CD= 5 2 miles through D,

and parallel to A4,draw DF

miles through F, andparallel to A5 , draw FG= 36miles ; and lastly, through

G, and,

parallel to A, draw

GH=58 miles. The point

H isx thfi .place the ship has

arrived at. Join AH , and through H draw HK parallel to

WE and meeting the meridian NS produced in K. Then AK

is the difference of latitude,and EH ,the departuremade goodduring the day. Also the angle KAH represents the direct

course, and the line AH the direct distance from A to H.

If we measure AK by the s cale of equal parts, we shall

find the difference of latitude AKabout 86milesto the south,and the departure KH about 42 miles

,to the east ofthe place

the ship left. The latitude arrived at is found thus

50°48

'N.

1 26 S.

49 22 N. and dep. 42'E .

By Trigonometry.

The diff. lat. and departure for each course and distancemay be computed as in Example p. 99. But to avoid this

tedious operation , the several quantities may be taken out of

the Traverse Table by inspection, as follows

104 PROBLEMS IN NAVIGATION.

log. AK

sec. KAH — 10

AH

distance, orAH= 95'8 miles.

A Ship from latitude 50° 48’s. has sailed duringthe

day on the following courses

1 . N.W . 20miles. 4. 15 miles.

2 . SW. 14 5 . N.N.W 18

3. 26 6. 29

Required latitude in , and departure ; and the direct courseand distance from one place to the other.

Ana. Latitude in 50° 5 ’

S. departure 209 miles ; course,N. 25

° 5 5’ W. dist. 47°8 miles.

The two following examples, taken out of that valuable

old work, Robertson’

s E lements ofNavigation, are given for

practice in construction, and for the Singularity of the form

of the diagrams resulting from the several courses and distances.

13. A ship sails from a place in latitude 40° N. on the

following courses. Required the latitude arrived at (by Construction).

29'

N .N.E. 10

E S E 50

E.N.E. 50

10

N.E .b.N 29

West 25 E QS.

S.S.E. 10

42 W.NW 31,~W

North 1 10 N.N.E.

EQN. 62 West

North 7

Am. The ship returns to the place sailed from.

PROBLEMS IN N'

AVIGATION. 1O5

14. Two Ships,Aand B,

part company in lat. 31° 31’N. ,

and meet together again at the end of two days, having run

as follows

The ship A.

Course.

1 .

2 . W .S.W.

3. E S E . W.S .W.

4. N.N.W. N.N.E.

Required the latitude arrived at, and the direct course anddistance of each Ship (by Construction) .Ans. Direct course due north

,and. distance 104 miles.

Lat. in, 33°15

’ N.

PROBLEMS IN SURVEYING,ASTRONOMY,

AND

NAVIGATION.

The following problems are given as Exercises in the pre

ceding rules of Plane and Spherical Trigonometry. They arealso designed to form an Introduction to Navigation and

Marine Surveying.

(Fig. 17 On the opposite bank of a river to thaton which I stood, is a tower known to be 2 16 feet high ; witha pocket sextant I ascertained the angle between ahorizontalline drawn from my eye (supposed to be 5 feet above the

ground) and its top to be 47°

required the distanceacross the river

,from the place where I stood to the bottom

of the tower. Ans. 1904 feet.

(Fig. BDC is a straight line to which AD is

perpendicular ; AD is 100 feet high, and subtends an angle

at B=36° and at C an angle=5 4°30

'

find the lengthof the line BC. Ans. 205 feet .

(Fig. A field is in the form of a right-angledtriangle, whose base is 200feet, and the angle at the base is67° how long will aman be walking round it at the rate of

4 miles an hour? Ans. 3 min. 2 l °6 sec.

(Fig. C and B are two places 100feet apart,and A is a point equally distant from C and B what

.

mustbe the distance of the point A from C or B that the angle

BAC may be Ans. feet.

(Fig. A Maypole being broken off by thewind, its top struck the ground at 15 feet distance from the

foot of the pole required the height of the whole Maypole,supposing the length of the broken piece to be 39 feet.

Ans. 75 feet.

The figures or diagrams for the following problems are not

drawn accurately to scale they are given merely to indicate the form,

without any regard to the value, of the sides and angles in the pro

blems to which they refer.


a horizontal plane, so that a plane passing through the three

points, A,C, and D ,

is perpendicular to the horizon ; the dis

tance CD= 100 yards, and the angle ACB=4G°

and

BDA=31° required theheight AB. Ans. 145 9 yards.

2d, When

’

CD=300yards,ACB=58°, and BDA=32

°.

Ans. 3074 yards.

(Fig. From the decks of two ships at D and C,880yards asunder, the angle of elevation of a cloud at A on

the same point of the compass from each is observed at Dthe angle is at C it is required the height Of thecloud above the surface of the sea, the

°deck of each ship beingsupposed to be elevated above it 2 1 feet. Ans. 942 6 yards.

A tower subtended 39° to an observer stationed200feet from the base find the height, and also the angle

it will subtend to an observer at 350feet from the base.

Ans. Height, 162 feet ; angle, 24°

(Fig. To determine nearly the distance between two ships at sea

,I carefully Observed the interval Of

time between the flash and report of a gun from each, and

measured the angle which the two ships subtended. The ln

tervalawere 4 seconds and 6 seconds, and the angle observed

48° required the distance of the ships from each other

(the velocity of sound being supposed to be 1 142 feet in a

second). Ans . Distance feet.2d

,When the intervals are 10secondsand 5 seconds

,and

Observed angle Ans. Dist . feet.

(Fig. From B, the top of a ship’

s mast,which

Was 80feet above the water, the angle of depression’of eu

other ship’

s hull at C upon thewaterwas required the

distance of the ships. Ans. 2 198 feet.

The angle between a horizontal line passing through the eye ofthe observer and a line drawn to the object is called the angle of els

ration when the object is above the horizontal line, and the angle ofdepression when the Object is below it ; thus in fig. BOA is the

angle of elevation of B above the horizontal line AO, and BBC is the

angle of depression of0below the horizontal line BD.

PROBLEMS IN SURVEYING. l09

2d, If the mast= 143 feet, and the angle of depression35

° Ans. feet.

[17] There are two monuments whose heights are 100

feet and 50feet respectively : I observed,with a sextant, thatthe line joining their tOps when produced made with the horizontal plane an angle=37

°required their distance apart.

Ans. 664 feet.

(Fig. Wishing to determine the height of anObelisk standing on the top of a declivity

,I measured from

its base a distance of 40 feet, and then observed the angle

formed by this line and a line drawn to the top to be=41°

.

Going on in the same direction 60 feet farther, the angle

formed by the declivity with a line drawn to the top was

23° Required the height of the obelisk.

Ans. Height, feet.

(Fig. To determine the distance between twoinaccessible objects, C and D, as two ships at anchor, a base,AB, on the same plane as the objects was measured andfound to be 670 yards ; the following angles were also oh

served at the extremities of the base at A,BAD=40°

BAC=97° 56'

at B, ABC=42

°and ABD= 1 13°

required CD . Ans. CD= 1 174 '4 yards.

(Fig. To determine nearly the distance of tworedoubts, C and D, by which the entrance into a harbour isdefended, a boat is placed at Awith its head towards a distant object seen at E, and the angles CAD=22

°and

DAE=48° 1'were Observed. The boat is then moved to B,

a distance of 1000yards, directly towards E,and the angles

CBD= 53° and DBE=75 ° 43' are observed ; requiredthe distance CD . Ans. Distance

, 1290yards.

A right-angled triangle rests on its hypothenuse,the length ofwhich is 100feet ; one of the angles is 36

°

and the inclination of the triangle to the horizon is find

the height of the right angle above the ground.

Ans. feet.

1 10 PROBLEMS IN SURVEYING .

(Fig. To determine the height of an object,EB

,on the top of an inaccessible hill, I took the angle of

elevation, ACE,of the top of the hill and also that of

the top of the object ACB=5 1°. Going then 100 yards in

a direct line from the object and upon a horizontal plane, Ifound the angle Of elevation of the top of the object ADB— 33° required the height of the object.

Ans. Height, yards.

(Fig. Wanting to know the distance between

two objects,A and B,which could only be seen from a part i

cular place, D ,I set up two staffs

,at C and E,

and took the

angles'

ADC=89°, ADB=72 and BDE=54° I

then measured DE and DC, each=200yards, and took the

angles BED=88° 30

’

and DCA=50° 30’

required the dis

tance AB. Ans. yards.

(Fig. To determine my distance from an in

accessible object at 0,without observing any angles, I mea

sured a straight line AB=500yards, from each extremity ofwhich I could see 0. I then measured fromA and B

,in a

direct line from 0,AC and BD, each= 100yards ; finally, I

measured the diagonals AI) and BC, the former was=5 50

yards, the latter=560 yards ; required the distance of the

Object from Aand B. Ans. AO=5 36°2,BO= 500°47 yds.

(Fig. Wanting to know the distance,AC , of

ahill from the station ,A, and also its height, CC, I measured

a base,AB= 298 yards, on ground nearly level, and at A and

B observed with a sextant the angles BAO=42°17

'and

ABO=79° 29’

and at A,the angle ofelevation OAC=4

°

required the distanceAC and QC .

Ans. AO=343°4, OC=29°14.

(Fig. Find how much the point B is elevated

above a point H,from the following data : having Observed

the angle of elevation BAH=4°

and measuredAC= 193

yards to a point C, at C the angle ACB=7G°

and at A

the angle BAC=45° Ans. BH= 17°8 yards.

1 12 PROBLEMS IN SURVEYING.

of the eye at each observation was 5 feet : find the height of

the hill above the lower end of the base. Am. 1226 ft.

(Fig. The distance between two stations B and

C on a declivity is 220yards. At B, the horizontal angle

Cba, between C and an Object A on‘

the top of a hill, was

found by a theodolzte to be 70° and at C, the horizontal

angle bCa, between B andA, was 62° the vertical angles

ACa=32° 12' and ECb=8° find the horizontal distances

of the Object A from C and B, namely Ca, and ba, and also

the heightsAa andAD of the object above C and B.

Ans. Hor. dist. ofA from C= Ca==279 °1 yards.

B ba=263°1

perp. height OfAabove C= Aa= 175 °7

B=AD=143°1

(Fig. Suppose, as in the last problem, thatthe direct distancebetween the stationsB and C Is 220yards,also that at C the angle BCb (the elevation of B)=8

° 32’

and ACa (the elevation of A)=32° but suppose now

that a sextant is used in measuring the oblique angles, and

that ABC=77° 8’

and ACB=62° Find, as before, the

horizontal distances between Aand the stationsB, C, and the

heights ofA above B and C . Ans. See last problem.

(Fig. Wanting to know the distance of two

objects, Aand B, from each other, and from another Object,D

, all in the same plane, on BA,produced on the side ofA,

a point C was taken, and CD being measured was found to

be yards, and the angle C=57°

. At D the angle

CDAwas observed to be and the angle BDA=41°

required the distance ofA, B,and D from each other.

Ans. AB=349°5 2,AD=4S7°27, and BD=498

°7 yards.

(Fig. Wishing to find the distance of abat

tery at B from a fort at F, which cannot be seen from the

battery in consequence of the ground between B and F beingcovered with °wood, &c.

, I measured distances, BA and AC,to points A and C

,where both the fort and battery were

visible, the former being 2000 yards and the latter 3000;

PROBLEMS IN SURVEYING. 1 13

and observed the angles BAF=34° FAC=74° and

FCA=80° From these data it is required to find the

distance of the fort from the battery. Ans . 5422 yards.

[37] (Fig. Coming from sea, at the point D , I Ob

served two headlands Aand B, and inland at C a steeple

which appeared between the headlands I found from amap

that the headlandswere 5 °35 miles from each other ; that°tha

distance fromA to the steeple was 28 miles, and from B tothe steeple miles. I Observed with a sextant the angle

ADC=12° 15 ’

and BDC= 15 ° required my distancefrom each of the three objects. (See Example, p .

Ans. AD= 1 1 '26, CD= 12’46, BD= 1 1

’ 03 miles.

(Fig. Required the distance of the three Oh

jectsA,B, and C from the point D,

situated within the triangle, from the following data AE=2G7, AC=346, BC=209

,anglesADC=128

°and ADB=91° (See Ex

ample, p. Ans . AD=249, BD=9 1‘

, CD= 131 .The following twelve problems require aknowledge of the

compass (p.

(Fig. A headland C bore due north of a shipat A; after sailing 10miles due east to B, the headland boreN.W. required the distance Of the headland at each Observa

tion. Ans. 10and miles.

(Fig. A fortAbore from a Ship C due north,by compass, and after sailingN.E. by compass 14; miles to B,

the fort bore N.W. find the distance of the fort at both ohservations. Ans. 205 and 145 miles.

(Fig. Aboat is placed atAdue west ofa ship

atB ; aftersailingN.W. 10miles to C, the boat bearsrequired the distance of the boat from the ship at the first

Ans. 10miles.

(Fig. Sailingalong a coast, a headland C wasobserved to bearN.E.b.N having run E .b.N. 15 miles toB,

the headland boreW.N.W. find the distance from the head

land at each observation. Ans. miles.

Aand B are two points lying north and south 50


yards apart ; what must be the distance of a third point Cfrom each, that it may bearN.b.W . from B, andW.b.S. from

A? Ans. and yards.

(Fig. A cape C was observed to bear from us

N.W . , and anotherheadland H to bear N.N.E .QE. standingaway 23miles to B,

we found the first bore fromus and the second N required the distance and bearing of the cape from the headland.

Ans. S. 5}W. ,miles.

A Ship sailing N.W.,two islands appeared in sight,

one bearing the otherN. from the Ship,and when

the ship had sailed six miles farther, the first boreand the otherN.E. ; required their bearing and distance from

each other. Ans. S . 58° 40’W .

, distance miles.

From two stations on the deck of a ship 100feet

apart, thebearingsofan object on ShorewereNE. andN N.E .

and the ship’

s head was N.b .W. find the distances of the

object from each station . Ans. 145 2 and feet.

[47] (Fig. A church C bears from a battery B,

E.N.E. 960yards how must the church bear from a ship at

sea,supposing her to run in until the battery is north 2000

yards ? Ans. N. 20°32

’E.

(Fig. A cape C bears from a headland H,

W.48. miles; howmust the cape bearfrom a Ship which

runs in towards the headland on a N.b.W.QW course, until

the headland is miles distant from the ship ?

Ans. W.N.W.

(Fig. A ship S was 2640yards due south of

a lighthouseAB , and after sailing 8001 ards toj ) ,

its angle of elevation BDAwas 5°

required its height .

Am . 192 yards.

(Fig. The bearings of two objects A and B in

the same latitude from a ship at C are N.N.E. andN

and the distance from A is 10miles ; required the distancefrom B. Ans. miles.

(Fig. Find the angle which the line of metal

1 16 PROBLEMS IN SURVEYING .

be 40, and the other two sides 20and 30,what is the length

of the perpendicular from the vertical angle ? Ans.

(Fig. The elevation of a spire DC at one station A was 23

°50

’

and the horizontal angle at this station, between the spire and another station B,was 93

° 4’

the horizontal angle at B was 5 4°28

’

and the distanceAB between the two stations was 416 feet required the

height of the spire. Ans. feet.

[58a] . (Fig. An observer finds the angle of eleva

tion of atower at a point B to be 23° 18’afterwalking from

B 300feet in a direction at right angles to the line joining B

with the foot of the tower, the angle of elevation was 2 1°

required the height of the tower, and its distance from B.

Ans. Height, feet ; dis’tance,

Apole is inclined at an angle of80°to the horizon

.tal plane, andwhen standing 100feet from it in the direction

in which it is incli e angle of elevation of the top is

54°

required its Ans. 1 12 5 feet.

[6os] . (Fig. 5 he bankof a river stands acolumn

200feet high, on which is a statue 50 feet high, and to an

observer on the opposite bank the statue subtended an equal

angle with a man 6 feet high standing at the base of the

column required the width of the river. Ans. 825 feet.

[61a] . (Fig. The distance between two objects, C

and D,is known to be 65 94 yards on one side of the line

CD there are two stations,Aand B,at which angles are taken .

The angle CAD=85° DAB=23° CBD=68° and

CBA=31° From these observations it is required to

find the distance between the stations A and B.

Ans. AB=4694 yards.

[62a] . (Fig. Aflags taff, 12 feet high, on the top of

a tower, subtended an angle of 48’

20”to an observer at the

distance of 100yards from the foot of the tower ; required

the height of the tower. Ans. 401 °4 feet.

PROBLEMS IN SURVEYING . l 17

Find the area of a triangle whose base is 40feet,and perpendicular 30feet. Ans. 66§ square yards.

[64a] . If from a right-angled triangle whose base is 12and perpendicular 16 feet, a line be drawn parallel to theper

pendicular, cutting off a triangle whose area is 24 feet,re

quired the Sides of the triangle. Ans. 6, 8, 10feet.

[65a] Required the side of an equilateral triangle, thearea of which is 180square yards. Ans. yards.

[66a] . Given the base of a triangle equal to 47 yards,

and the angles at the base 27° 10

’

15"and 35

°10

'

to

find the area. Am. 33680.

[67a] . The area ofa triangle is 6, and two of its sides are3 and 5 find the third side. Ans. 4 or J 5 2 .

[68a] . (Fig. The straight line EFis drawn parallelto the base of the triangle ABC,

whose altitude is 10feetfind the distance of EF from the base BC, so that itmay divide the triangle into two equal pa

rts.

Ans. Distance from baSe, 2 °929 feet.

[69a] . When a parish was enclosed, the allotment of oneof the proprietors consisted of two pieces of ground, one of

which was in the form of a right-angled triangle the other

was a rectangle, one of the sides of which was equal to thehypothenuse of the triangle, the other to half the greater sidebut wishing to have his land in one piece, he exchanged hisallotments for a square piece of ground ofequal area, one sideofwhich equalled the greater of the sides of the trianglewhich

contained the right angle. By the exchange,he found he hadsaved 5 5 yards ofpaling. Required the area of the triangled rectangle. Ans. Triangle, rectangle,

[7ou] . The areaof a right-angled triangle whose sides arein arithmetical progression : 216 determine the sides.

Ans. 18, 24, 30

[71a] . What Is the side ofthat equilateral triangle whosearea cost as much paving, at 8d. per foot, as the palisadingthe three Sides did at 78. a foot ? Ans. 72 74 feet.


[72a] . Given a side, a, of a regular polygon of n sides, to

find a formula for computing the area.

EXAMPLES.

1 . An eight sided polygon, or octagon, whose side=16

yards. Ans. Area= 1236°1 .

2 . A ten-sided polygon , or decagon, whose side=20§~

yards. Ans. Area=3233°5 .

[73a] . Given the areaA of a regular polygon of n sides,to find a side a.

Ans. a

ExAMPLES.

1 . The area of a regular octagon is 1236 1 square yardsfind a side. Ans. Side 16 yards.

2 The area of a decagon is 3233°5 square yards ; find a

Ans. Side 20°5 yards.

[74a] . (Fig. To make a regular polygon Of n sides

equal to a given triangle ABC . Take AD 953; draw AG,

360°

making angle CAGndraw DE parallel to AC ; and

take AF=a mean proportional to AC and AR thenAF isthe radius ofa circle that will contain the required polygon

required a proof.

[7 A ship sailing on a S S.W. course, bore from medue south, and the angle subtended by the head and stemwas 20

’and her length was known to be 160feet ; re

quired her distance. Ans. 1 °94 miles.

[76a] . A ship sailed S. and met anothership which

had sailed N 1O°)W. from the same meridian ; the dis

tances sailed were as 3 2,and their distance from the me

ridian left was 100miles; required the difference of latitude.

Ans. 47 miles.


[86a] . In a plane triangle ABC, given A=SO°

, and a

400 required the other sides,their sum being 600.

Ans.

[87a] . In a plane triangle ABC, given A=6O°

, c=400,and the sum ofthe other two sides, a and 6, =600 required

the Sides a and b. Ans. 250and 350.

[88a] . The perimeter of a right-angled triangle is 24 feet,and its base is 8 feet ; find the other sides.

Ans. 6 and 10feet.

[89a] . At 80feet distance from a steeple, the angle madeby a line drawn from its top to the place was double thatmade by a line drawn from the top to a point 250feet fromthe steeple on the same level ; required the height of thesteeple. Ans. 150feet .

[90a] . Given the base a, the vertical angle A,and the

sum of the other two Sides of the plane triangle ABC=b ; tofind the sidesa: and g.

Ans. x+y=b, and- a)

fromwhich equations asand 31 may be found.

[91a] . Given the base a, the vertical angle A, and the

difference ofthe sides ofa plane triangle:-

,d to find the sides

s and y.

Ans. z y=d, and — d) cosec fiéz

[92a] . Given the base a, the difference of the angles a:

and g at the base=d, and the sum of the two other sides of

the plane triangle : b, to solve the triangle.

Ans. Cos. s-cos.

21 d

, which determines the sum

of the unknown angles, and thence,with their difi'

er

ence already known,the angles a: and g.

[93a] . Given the base a, the difference of the angles a:

and g at the base=D,and the difference of the Sides of a

plane triangle : : d, to solve the triangle.

Ans. Sin. and :c—y is alreadyknown .

PROBLEMS IN l UBVEYING.

’

121

[94a] . Given the base of a triangle b, and one of the

angles at the base A,and the difference ofthe other sides d,

to solve the triangle.

b dAns. Tan . 40

b d

found, and thence the other parts of the triangle.

[95a] . Given the base of a plane triangle b, one of the

angles at the base .A,and the sum of the other sidesz m, to

solve the triangle.

tan QA, from which the angle C is

m b

m+ bcoté A.

[96a] . Given the angles and the perimeter of a planetriangle, to find the sides.

Ans. Tan . QC

EXAMPLE.

Perimeter=100yards,A= 102° 5 1 ’

B=25 ° 42'45 and

C=5 1° 25'45 AM . a=44 '5 1 , b= 19

°8, c=35 '69 yards.

[97a] . Given thedistances between three stations in a

straight line with an object standing upon a horizontal plane ;and the angles at the points E

,D

, C ,=0, 90— 0

,and 20in

order, 0being unknown,to find its height.

EXAMPLE.

Let ED=20,and DC=20. Ans;

[98a] . Investigate analytical expressions for calculatingthe distance of a station from each of three points, havinggiven the distances of the points from each other, and the

angles which they subtend at the station . For numerical

examples, see Probb. 37,38.

[99a] . If in the three edges which meet at one angle of acube, three poin ts A,B, C, be taken at distancesa, b, c, from

the angle respectively, the areaof the triangleABC,formed by

joiningthethree pointswith each other, J a% 3 a§lz9 b’cz.

[l00a] . On the sides Of an equilateral triangle three

squares are described . Show that the area of the triangle

formed by joining the centres of these squares=area equila

teral triangle x (1


[lOla] . A fleet of steam-vessels, under the command of

an admiraLissteamingdue east, 6milesan hour ; apart ofthefleet under a Vice-admiral bears from the admiral N.N.E. 6

miles, and a look-out steamer is 10miles. The ad

miral, suspecting the enemy to be in the S E. quarter, directsthe vice-admiral

,by signal, to take station 3 miles SE. of

him, and the look-out steamer due east 5 miles at the same

time he sends several steamers from the main body'

to rein

force the Vice-admiral at the appointed station . Now, sup

posing the admiralcontinue to steerdue east 6miles an hour,and the vice-admiral to increase his speed to 10miles, andthe look-out steamerto 14 milesan hour, it is required tofind

the course and distance of the look-out steamer, and also thecourse and distance of the vice-admiral, so that they inayreach their several stations in the shortest time. Requiredalso the time in reaching their stations, the rate at which thereinforcement must steam to reach the station at the same

time as the vice-admiral, and also its course and distance.Ans. Viceiadmiral’s course is nearly ; distance,

miles. Look-out steamer’

s course is NlN.E .

nearly ; distance, miles. Reinforcement’s courseis nearly distance, 8 miles. The Vice

admiral and reinforcement reach their station in 56minutes, and the look-out steamer in 38 minutes.

The reinforcement’s rate of steaming, 8§ miles an

If a geometrical construction of this problem be made on

a large scale, the course, &c. so found will be sufficiently cor

rect for practical purposes. (See a geometrical constructionin the volume of Solutions. )

The above example belongs to a very important class of

problems connected with naval tactics and the combined evolutions of steam squadrons. These problemswillattract more

attention hereafter, when .it is found that a fleet of steamers

must be handled with almost the same precision as an army.

124 DEFINITIONS IN ASTRONOMY.

duce it to the celestial concave at M then M is called theapparent place Of the heavenly body m.

The extremities Of the axis on which the celestial

concave appears to revolve, in consequence Of the earth’s

diurnal motion about its axispp l , are called the poles of the

heavens . thus the line PP represents the axis Ofthe heavens,P and P are the poles ; the points p and 391 in which the

earth’

s surfacepqplql is cut by this axis, are called the poles

of the earth. That great circle on the surface Of the earthwhich is equidistant from each of its poles is called the terres

trial equator ; ih ths’

figure, ggl represents the plane Of the

terrestrial equator. The terrestrial equator extended to thecelestial concave, as QQI, forms the celestial equator. A circle

touching the earth where the spectator stands, and extendingto the celestial concave, is called the visible horizon , and a

circle parallel to the visible horizon which passes through thecentre of the earth and extends to the celestial concave is

called the rational horizon ; thus hr represents the visible,and HR the rational horizon Of the spectator at A. These

two circles, however, form one and the same in the celestial

concave thus the points R and r in the figure must be supposed to coincide. Thismay be readily conceived, when weconsider that the distance Ofany two points on the surface ofthe earth will make no sensible angle at the celestial concavetherefore either Of these two circles is to be understood by theword horizon. Of the poles Of the horizon Of any place, that

which is over the place is called the zenith, and the otherthe

nadir, as Z and z, in fig. a.

Great circles passing through the zenith are called

circles of altitude, or vertical circles the circle Of altitudepassing through the poles Of the heavens is also called the

celestial meridian the points Of the horizon through which

the celestial meridian passes are called the north and southpoints and a circle Of altitude at right anglesto themeridianis called theprimevertical. The points Of the horizon throughwhichtheprimeverticalpassesarecalledtheeastendwestpoints.

DEFINITIONS IN ASTBONOIIr. 1 25

These circles will be more clearly understood bymeans of fig. 6

,in which NWSE represents the horizon of a

spectatorwhose zenith is supposed to be Z ; N, S, E,W the

north, south, east, and

west points Ofthe horizon.

NZS is acircle ofaltitudepassing through the zenithand pole, and therefore represents the celestial meridian : ZD, 20, &c. are

circles Of altitude, Or ver

the prime verticalSince the horizon

and celestial equator are

both perpendicularto the celestial meridian , the points wherethe horizon and celestial equator intersect each othermust be90° distant from every part of the meridian [Part II. Art .

Cor. 3] that is, the celestial equator cuts the horizon in

the east and west points draw the curve EQW to cut the

horizon in the east . and west points ; this will represent thecelestial equator. Since the poles Of the heavens are 90° distant from the equator, take QP= 90

°. Then P is one ofthe

poles of the heavens ; it, is called the elevated pole, or theone above the horizon

,and the arc NP is the altitude Of the

pole.

The distance, ZQ (figures a and b), Of the zenithfrom the equator represents the latitude of the spectator thismay be more clearly seen in fig. a, p. 123, where PZQH re

presents the plane Of the celestial meridian, Z the zenith ofthe spectator at A, hr or HR his horizon, P the elevatedpole, QQI, drawn at right angles to P1P, represents the planeOf the celestial equator

,and the angle ZGQ (the earth being

considered as a spheroid) is the latitude of the spectator ; butSince the distance, ZQ, Of the zenith from the equator mea

sures the same angle, this are is taken to represent the lati


tude.

* ZP (figures a and b), the complement Of ZQ, is called

the colatitude.

The altitude Of the pole is easily Shown to be equalto the latitude Of the spectator ;

for alt. NP=90° — PZ=90° — colat._ 90° (90

°

The sun,in consequence Of the earth’

s motion inits orbit, appears to move eastward in the celestial concave,and, in the course of a year, to describe among the fixed starsagreat circle ; this circle is called the ecliptic. It is inclinedto the celestial equator at an angle Of about 23° called

the obliquity Of the ecliptic. In fig. 6,ACT represents a partOf the ecliptic, and the angle MAR the obliquity. The

points in which the equator and ecliptic intersect are called

thefirstpoint of Aries and first point of Libra ; the former,being the pointwhere the sun crosses the equator to the north

ward, is called the vernal equinoctialpoint, and the lattertheautumnal equinox. Great circles passing through the polesOfthe heavens are called circles of declination, and great circlespassing through the poles ofthe eclipticare circles Of latitude:thus, in fig. b

,PR

,PA, &c. are circles Of declination ; and if

Plrepresent the pole of the ecliptic, PIM is a circle of lati

tude. Parallels of declination and latitude are small circles

parallel respectively to the celestial equator and ecliptic.

The declination and right ascension Of a heavenly

bodymay be defined thus the declination Ofa heavenly body

is the arc Of the circle Of declination passing through its placein the celestial concave, intercepted between this place andthe celestial equator ; the right ascension Of a heavenly bodyis the arc Ofthe celestial equator intercepted between the first

point ofAries and the circle Of declination passing throughthe place Of the body in the celestial concave. Or, it is the

If a line be drawn from C,the centre Of the earth, to A, and

produced to cut the celestial concave in Z“ the are Z,Q is called the

latitude on the sphere, or the reduced latitude.


Definitions in Navigation .

Let Aand F represent two places on the surface of

the earth (considered as asphere) ,PU, PZ,

their meridians, P the

pole, and UZ an arc of the equator. Through A and F draw a

curve line,AF, cutting all the

intermediate meridians,PV

,PW

,

&c. at the same angle. Thiscommon angle is called the course

fromA to F, and the arc AF (innautical miles) is called the distance. Draw the parallels of latitude AA’

and FF’

; the arc AU

is the latitude OfA,and FZ the

latitude Of F ; UZ, or the angle

APF, is the'

difference Of longi

tude between A and F . The are AF' is the difference, or as

it is called in Navigation, the true diference of latitude be

tween A and F.

Suppose the intermediate meridians, PV , PW,&c.

to be drawn through‘

points B,C,&c. taken on the arc AF

indefinitely near to one another, and thoughA,B,&c. suppose

arcs OfparallelsAM,BN, &c. tobe drawn ; on this supposition

the elementary triangles ABM,BCN,

&c. may be consideredas right-angled p lane triangles, of which the sum ofthe sides

AB,BC, &c. is the distance, the sum Of the Sides BM

,NC

,&c.

is equal to the true difference of latitude, and the sum Of the

sides AM,BN

, &c. is called in Navigation the departure.

The chart used at sea for marking down the ship’

s track

and for other purposes,exhibits the surface of the globe on

a plane,on which the meridians are drawn parallel to each

other,and therefore the parts AM,

BN, CO,

&c.,arcs Of par

allels Of latitude, are increased and become equal to the cor

responding parts of the equatorW ,W ,&c. Now, in order

RRORLmIS IN NAUTICAL ASTRONOMY. 129

that every point on this plans may occupy the same relativeposition with respect to each other that the points corresponding to them do on the surface Of the globe, the distance

between any points A and F’

, and A and F, must be ih

creased ih the same proportion as the distance FF’has been

increased . The true difference Of latitude AF’

,is thus pro

jected on the chart into what is called the meridional difer

ence of latitude,and the departure AM+BN+CO ia

to the difference Of longitude. A chart constructed in this

manner is called aMercator’

s Chart. From these‘

definitions

and principles are deduced the fundamental formula in

Navigation ; and those, expressed in words, form the commonrules Of Mercator and parallel sailing. The investigation of

these formulae and mles is given in Navigation, Part II.

Spherical Trigonometry derived its origin from the com

putations necessary in astronomy,and its principal applica

tions are still furnished by the same science.

Examp les in NauticalAstronomy.

1 . In latitude 30° 45 'N. the altitude of the sun was

14°45

’

(east ofmeridian), and by the NauticalAlmanac its

declination at the time Of Observation was 16° 10’

8 ; required

the azimuth and hour angle.

Let NWSE represent the horizon, and Z the zenith ofthe

spectator ; NZS the celestial me

ridian : then WZE, drawn‘

at

right angles to themeridian, willrepresent the prime vertical (seedefinitions, p . 1 Take NP

an arc of the meridian=30° 45 ’

w

the latitude ; then P is the poleof the heavens

,since the altitude

Of the pole is equal to the lati

130 PROBLEMS IN NAUTICAL ASTRONOMY.

the west and east points Of the horizon, andQ, draw the circleWQE ; this will represent the celestial equator.

LetX be the place of the sun at the time of Observation .

Through X draw ZO a circle of altitude, and PX a circle of

declination then PZX is a Spherical triangle, in which are

given the polar distance PX=90° declination=106°

the zenith distance ZX=90°— altitude=75 ° and the cc

latitude PZ=90° — latitude=59° 15 ' to find the anglePZX

the aziniuth, and the angle ZPX the hour angle Of the hea

venly body.

Calculation. (By Rule VIII.)

(1 To find azimuth PZX, or are NO .

75 ° 15’

0014553

5 9 15 0065801

16 0 49 42169

106 10 48 501 15

122 10 2638

90 10 PZX=1 19° 27'

or azimuth=N. 1 19°27

’

E.

TO find hour angle ZPX. (By Rule VIII. )106

°10 00175 23

5 9 15 0065801

46 5 5 49 42169

75 15 43 8871 1

122 10 9 414204

28 20 ZPX=4h. 5m. IS. the hour anglea

The place Of the heavenly body X in this, and in all the dia

grams in the following pages, should be estimated as near the truth

as possible by the eye without using mathematical instruments ; a

tolerably correct figure may be always drawn in this manner after alittle practice. For instance, Since the above figure is projected fora place in north latitude,X must be placed below the equatorbecauseits declination is south ; the altitude X0 in

.

the figure is aboutand declination RX about supposing the lines Z0 and PR torepresent


3. Required the right ascension and declination of the

sun when its longitude is 64° 33

’the obliquity Of the

ecliptic being 23° 27

’

LetAR represent the celestial equator,andAX the ecliptic;the intersection A

the first point Of

Aries. Let X be

the place Ofthe sun

in theecliptic: then

64°33

'

and the angleXAR=the obliquity Of the ecliptic— 23° 27

’

Through X draw XR, a circle ofdeclination

then XAR is a right-angled‘Spherical triangle, in which are

given AX, the angleA, and the right angleARX to findXR

the sun’s declination , and AR its right ascension (see defini

tions,p.

Calculation. (By Rule XIII .)

To find right ascension AR. TO find decl. XR.

Sin . XR=Sin. A Sin . AX

cos. A=cot. AX. tan. AR

AX 64°

33'15

"199 625 21 9 9 5 5 684

23 27 45 7439 9 5 5 5 729

102 85082 XR=2 1° 4’

15 N

AR=4h . 10m. 208. the sun’

s declination.

the sun’s right ascension.

4. Required the time Of sun-rising at a place in latitude30° 45

'S. ,

the sun’s declination being 18

° 28’ N. (neglecting

the effects of refraction,Let NWSE represent the horizon

,NZS the celestial me

ridian . Then, Since the Spectator, whose zenith is Z, is in

south latitude, the south pole is above the horizon : take,

therefore, SF=3O°45

’

the latitude ; then P is the south pole .

Let 0 be the place of the sun in the horizon, and through 0

PROBLEMS IN NAUTICAL ASTRONOMY.

draw the circle Of altitude 20 and declination

then ZPO is a quadrantal tri

angle, in which are given PZ

the colatitude 5 9° 1 the

polar distance PO= 90° 4» declination 108

°and the

quadrant ZO=9O° to find the

angle ZPO,the hour angle at

the time Of rising, which sub

tracted from 12h. will give theA.M. time Of sun -rising.

To find.

hour angle ZPO. cos. P= — cot. PZ cot. PO

59° 15’

9 774471 (By Rule XIV.)100 28 9 5 23679

P=5h. 14m. 108. 92 98150

sun rises at 6h. 45m. 508. A.M.

Trigonometry. They also form an Introdu ction toNautical Astronomy.

[102] In latitude 50° '

48’ N. the altitude Of the sun was

46°20

’

(west Ofmeridian), and, bythe Nautical Almanac,its declination at the time of observation was 23° 27 45"N.

required the azimuth. (Fig. Ans. N. 1 1 1°5 1

'W.

In latitude 50° 48’ N. the altitude Of the sun was46

°20

’

(west ofmeridian), and its declination 23°27

’

45’ N. ;

required the apparent time Of Observation. (Fig. 7

Ans. 2h. 57m. 16s.

[1‘The azimuth Of a heavenly body was observed to

be N. 1 1 1° 5 1

’W. , its altitude at the same time was 46°

and declination 23° 27’

45"N. required the apparent time.

(Fig. Ans. 2h. 5 7m. 16s.

l34 PROBLEMS IN NAUTICAL ASTRONOMY.

What is the altitude Of a starwhose hour angle is38° and declination 16° N. at a place in latitude 50° 48'N

(Fig. Ans. 43°

In latitude 50° 48’ N. , when the sun's declination

was 12°29

’ N. and hour angle 2h. 53m. ls. A.M. requiredthe azimuth. (Fig. p . Ans. N. 121

°47

'E.

[1 Given the sun’s meridian altitude 70° (zenith

north Of the sun) and its declination 20° N. required the

latitude of the place . Ans. 40° N.

Given the sun’

s meridian altitude=70° (zenithnorth) and its declination 5

° S . ; required the latitude. (Fig.

Ans. 15° N.

[1 Given astar’smeridian altitude=70° (zenith south)

and declination 25 ° N. required the latitude. (Fig.

Ans. 5° N.

[1 1 Given the sun ’smeridian altitude 30° (zenith south)

and declination 10° N required the latitude. (Fig.

Ans. 50° S .

[1 1 Given the latitude ofthe spectator40° N. and the

declination ofa heavenly body 20° N required itsmeridian

altitude. Ans.

[1 12] Given the sun’

smeridian altitude 30° (zenith southof the sun) at a place in latitude 50

° S find its declination .

Ans. 10° N.

[1 1 The meridian altitude of a star at a place on the

equator is 57° find its declination (zenith north Of star).Ans. 33° 8.

[1 1 The latitude Ofa spectator is equal to half the sum

of the meridian altitudes of a circumpolar star aboveand be

low the pole required a proof. Given the altitudes 20° and

70° at its inferior and superior transits respectively ; required

the latitude. (Fig. Ans. 45 ° N.

“

The latitude and declination in the astronomical problems areto be considered north, unless the contrary is expressed.

136 PROBLEMS IN NAUTICAL ASTRONOMY .

3. When the heavenly body is in the horizon, as at D

008. hour angle - tan. lat. x tan. decl.x. Sin . decl.=cos. lat. x sin. amplitude.

A. Sin. lat. - cot. amplitude x cot. hour angle.

n. cos. amplitude=cos. decl. x Sin . hour angle.

Given the sun’

s altitude 30° when due west, and

its declination 20° N. required the latitude.

Ans. 43°

9'15 N.

Given the sun’s declination 23

°27

’45 and lati

tude of the place 50° 48’ N. to find his altitude, and the

time when he is on the prime vertical.Ans. Alt. 30° 5 5 ’ time

,4h. 37m. 48.

Two starsare due cast at the same time at a place

whose latitude is 50°

48’ N and their altitudes are 20° and

find the difference Of their hour angles.

Ans. 5 9m. 568.

Given theSun’s altitude 18° 45 'at Six O’clock, and

declination 20° 4’ N to find the latitude.

Ans. 69°

31’40 N.

Given the latitude 50° 48'N and sun’s decline?

tion'

23°

27'45

” N. to find his altitude and azimuth at.

Six

O’clock. Ans. Alt. 15”

azimuth, N. 74°

39’

30”E.

What will be the apparent time of rising of the

sun at a place in latitude 50°

48'N when itsamplitude is E .

10° S. (neglecting the effects of refraction,

Ans. 6h. 31m. 78.

Given the sun’samplitudeW. 37 30

’ N. and decl.15

°12

’ N. required the latitude. Ans. 64°

29’

15” N.

Given the latitude of the place 50° 48'N. and the

sun’s decl. 18° 28'N. required his amplitude, time Of rising

or setting, and the length of the day and night.

Ans. Amp. E. 30°4

’

30" N. sun rises at 4h. 23m. 19s.

A.M. length of day 15h. 13m. 223.

Wherewill thesun rise in latitude 50° 48’ N. when

the day is 14 hourslong ? Ans. E. 19°

5’N,

PROBLEMS IN NAUTICAL ASTRONOMY. 137

Given the sun’

s altitude 22° and hofIr angle

3h. 0m., when the declination is nothing, to find the latitude.

Ans. 56° 33

’30

” N.

Given the sun’

S altitude 42° declination,N. and azimuth

,S. 5 7

° 45"W. tofind the latitude.

Ans. Lat 5 9°4

’ N.

Given the sun’

s altitude 37° hour angle 2h.

1 5m. , and decl. 22° 30

’N. to find the latitude (zenith northof the sun) . (Fig. Ans. Lat. 71

° 31'N .

Given the right ascension of the sun (RA), 4h.

10m. 208. and obliquity of the ecliptic (to), 23° 27

’to

find his longitude and declination . (Fig.

Ans. Long. 64°33

'(1801. 2 1

°

4’

15 N.

2 . Let RA=l7h. 10m. ; w=23° 27

'

(Fig.

‘Ans. Long. 25 8° 30

'

15” decl. 22° 58’ S .

Given the right ascension and declination Of a

heavenly body, and the obliquity of the ecliptic to find its

latitude and longitude. (Figs. 77, 78

EXAMPLES.

1 . Let RA,=2h. 5 9m. decl.=21° 27'48 N. ; w

23°27

'45 Ans. Lat. 4

° 15'N. long. 48

°37

'30

2 . {Let RA=16h. 14m. ; decl.=25°5 1

’ N. ; w=23°

27’45 . Ans. Lat. 46

° 6’15 N . long. 234

°36

’

Two places have the same latitude namely, 45

°N.,

and their difference of longit ude is 10° 36

,required their

distance. (Fig. 79 . Ans. 449 25 nautical miles.

[ 1 Required the distance from Portsmouth to BuenosAyres. (Fig.

Lat. Portsmouth, 50° 48’N. long. 1° 6

’W.

Buenos Ayres, 34 37 S . long. 58 24 W.

Ans. 5949 -8nautical miles, or 6847-2 English miles

(697916 to a degree) .


Sailing on or near to a Great Circle.

The Shortest distance between two known places is

the arc ofa great circle passing through them. This distance

may be found as in the last example. The practical incon

venience of sailing on a great circle arises from the necessityof continually altering the course. It is for this reason thatRules are given for sailing from one place to another SO thatthe coursemay be constant, although thedistance sailed is not

the Shortest between the two places. When the distance be

tween the two places is considerable, as between Liverpool

andNew York, the followingmethod Ofapproximating to theShortest distance may be adopted with advantage

1 Compute the Shortest distance as in the last example.

2 . Take two ormore points on this arc, and find the lati

tude and longitude of those points.

“

3. Find the courseand distance from the place sailed from,

to the nearest point marked on the arc, and thence to the

next point, and SO on. (Navigation, Part I. )The distance described by the Ship on these curves will

not differ much from the Shortest distance and thus the

advantage of sailing on one course is combined .with that of

Shortening the distance. This will be seen in the followingexample :

Eaample in Great-circle Sailing.

1 . Conipute the Shortest distance between Liverpool andNew Yorkqt

Lat. Liverpool, 5 3° 25

’ N. long 2° 59’W.

New York, 40 42 N. ; 73 5 9 W.

Ans. 2872 miles.

The points Should not be farther apart than 1000miles. The

nearer they are taken to each other, the less will be the difierence

between the sum of the distances rou nd the Shortest distance.

1' This example is worked out at length in Navigation, Part II.

pp. 167-169.


C to the north of and 20° distant from Aand B. (Figs. 86,

Ana. Lat. N. long. 2 1°

13'E. ,

or 8°47

'E.

The distances Of a comet from Aldebaran and

Reguluswere observed to be 40° 12’

and 5 1°36

'

respectively ;required its latitude and longitude, the respective latitudes of

the two stars being 5°28

’

45” S . and 0

°27

’

30" N. ,

their

longitudes 67°

12’15 and 147

°1 5

'

and the comet beingsouth-east of the arc Of a great circle joining them. (Fig.

Ana. Lat. 28°

0’

15” S. long. 102°

Required the beginning ofmorning and the end Of

evening twilight at a place in latitude 5 4°

36'

N the sun’

s

declination being 8°30

'N. (twilight being supposed to begin

and end when the sun is 18° below the horizon). (Fig.

Ans. 2h . 46m. A.M.,9 h . 14m. P .M.

Suppose two altitudes Of the sun Observed in theforenoon in the same place, at

'

the interval Of an hour and a

half, to be 28°

40’

and 39°

50’

required the latitude of the

place,the declination being 23

°

26’ N. at both Observations.

(Fig. Ans. Lat. 5 9°

17’15

" N.

The altitudes of a Hydra and Regulus (both eastOfmeridian) were observed at the same time to be 40° 44’

and 45°respectively, the right ascension and declination of

the former being 9h. 16m. and 7°37

’

S. , Of the latter 9h .

53m. and 13°9

’ N. required the latitude. (Fig.

Ans. Lat. 26°

23’ N.

The difi‘

erence of longitude between twoplaces inthe same latitude, namely 33

°

5 1’ S. is 136

°

how much

Shorter is the distance between them on the arc of a great

circle than on their common parallel, the distance measured

on the parallel being 6785 miles ; and what is the highest

latitude attained by a Ship sailing from one to the other on

the arc of agreat circle ? (Fig.

Am . Difference of distances, 7376 nautical miles

highest lat. 60°

54’ S.


What is the highest latitude attained by a Shipsailing on the arc

'

Ofa great circle fromPort Jackson to Cape

Horn, their latitudes being 33°5 1

'

and 5 5°

and the differ

ence Of longitude 140°

(Fig. Am. 72°

Required the sun’

s azimuth and depression below

the horizon at 7h. P .M., when the declination was10

°1 5

'S. ,

and the latitude Of the place 50° 48’ N. (Fig.

Ans. Azimuth N. 84°

5 3’ W. depression , 17

°

Determine the bearing or azimuth of the two starsAldebaran and Pollux when on the same vertical circle, thelatitude Of place being 25

° N. the RA. and declination Of

former star being 4h . 26m. 468. and 16°

1 1'

N of the latter7h. 35m. 138. and 28

°

24'26 N. (Fig.

Ans. Bearing ofAldebaran , S . 75°5

’W.

In a certain latitude (zenith N.) the moon’

s truealtitude was 18° 2’

30”

(east ofmeridian) in the same vertical

circle with a star whose right ascension and declination were

9h. 5 9m. 42s. and 1 2°

45’

45” N the moon

’

s RA. and de

clination being 1 2h . 35m. 54s. and 1°42

’30

”S. required

the latitude. (Fig. Ans. 19°

55'30

” N.

The following problems are given as exercises.

in ana

lytical trigonometry.

[150a] . The altitude Ofa star when due cast was and

it rose E.b.N. required the latitude. Ans. 29°

42'N .

[15 1a] . The altitude Of a star when due east was

when due south was 40° fin d the latitude Of the place.

Ans. 58°

31'N.

[152a] . Given the altitude of the sun when due west and

at Six O’clock, to find the latitude and declination .

EXAMPLE.

Altitude when we8t=27° at 6h.= 14°

43’

30

Ans. Latitude, 48° N declination , 20

°

[153a] . Given the sun’

s altitude at Six O’clock, and

amplitude ; to find the latitude and declination.


Altitude at 6h.= 14° 43’

30 amplitude=30° 44’30

Ans. Latitude, 42°or declination, 22

°20

’

or

[154a] . Given the.

sun’

s altitude at Six O’clock, and the

hour angle when setting; to find the latitude and declination.

EXAMPLE.

Altitude at 6h.= 14° 43’

30”h=7h. 35m. 228.

Ans. Latitude, 48°

1,declination, 20

°

[155a]. Given the times at which the sun sets and 18 west

on the same day, at a particular place to find the latitude ofthe place, and the declination. a

Hour angle when W .=4h. 43m. 28s.

setting=7h. 35m. 228.

Am. Latitude, 48° N. ; declination, 20

° N.

[156a] . Given the sun’s declination , and the interval be

tween the times at which he is west and sets ; to find the

latitude.

EXAMPLE

Declination=20° N. ; interval=2h. 5 1m. 54s.

Ans. Latitude, 48° N. or 42

° N.

[l57a] . Given the amplitude of the sun ,and the azimuth

at Six 0’

clock ; to find the latitude and declination.

Amp] .=W . 30°

44'30

” N. azimuth=N. 76°

18’45

”W.

Ans. Latitude, 48 N declination, 20

° N. nearly.

[158a] . Given the sun’s meridian altitude, and his alti

tude at Six o’

clock ; to find the latitude and declination .

EXAMPLE.

Mer. alt. alt. at 611 .=14°

43'30

”

Ans. Latitude, 48°

0’20

” N. declination, 20° N.


time Of the Observation, the difference of their R.A.

’S being

1h. 5 3m. 428. (Fig.

Ans. App. time, 5h . 9m. 403. lat. 50°

18’

20” N.

[l64a] . On what days Of the year is the sun on the hori

zons of Dublin and Pernambuco at the same instant, their

respective latitudes being 53°

21’ N. and 8

°

13'S.

,and their

longitudes 6°

1 9’ W. and 35

°

5'W . ?

Am . The four days in the yearwhen the sun’

s de

clination is 18°

[165a] . ABCD‘

iS asquare field, each Ofwhose Sides is 100

yards. In the middle Of the field stands an Obelisk 60feethigh ; find the altitude of the sun when the shadow of the

Obelisk just reaches the corner Of the square. Ans. 15°

[166a] . At noon on the Shortest day the Shadow Ofa per

pendicular stick was 7 times as long as its shadow at noon on

the longest day ; required the latitude, the declination'

being

23°

(Fig. Ans. Sin . 2l=§ Sin . 2d ; l=38°

27’

[167a] . Compare the lengths Of the Shadow of a perpen

dicular stick at noon in latitude 45°N. on the two dayswhen

the sun’s declination was 15° N. and 1 5

° S.

Ans. Length of Shadows as 3 1 .

[168a] . In latitude 33°30

'N. and decl. 10° 1 5'N I oh

served that my shadow bore to my height the proportion Of

5 3 required the altitude and hour angle of the sun .

Ans. Alt. 30° 58'

h=3h. 58m. 4S .

[169a] . The length Of the Shadow Of a perpendicularObject was 4 feet, and its longest when sloping was 5 feet ;required the sun

’

s altitude. (Fig. Ans. 36°52

'I5

”

.

[170a] . The elevation Of a cloud was observed to beat the same time the sun ’

s altitude was the sun and cloudbeing in the same plane with the Observer, and his distancefrom the shadow 400yards; required the height Ofthe cloud.

(Fig. Ans. 1468 yards.

[17la] . In latitude 45° N, the meridian altitude of the


sun Show that the tangent Ofquarter the length Of the

1

~73° (Fig.

[172a] . At a certain place the sun rose at 7h. A.M. , and

itsmeridian zenith distance was twice the latitude ; requiredthe latitude. Ans. Lat. 26°

[173a] In latitude 45° N. the sun rose at 4h. A.M. Show

that the tangent of the meridian altitude=3.

[174a] . In latitude 50° N. when the sun

’

s declination is5 ° 38

’ N required the time it will take the body Of the sun'

to rise out Of the horizon , its semidiameter being (Fig.

Ans. 3m. 198.

[175a] . Required the time the sun’s semidiameter will

take to pass the meridian, the declination being 23°4

’and

semidiameter 16’

(Fig. Ans. 1m. 1 18 .

[l76a] . In latitude 5 9°6

’ N. and long. 6° 15

'E . Observed

a point of land bearingN.E.

, and after sailing E .N.E. 6miles

the point bore N lW required the latitude and longitudeof the point . (Fig.

Ans. Lat. 5 9° 1 1

’

long. 6° 25

’13

”

E.

[177C] . In latitude 45° N required the difference in the

lengthsof the longest and Shortest days (declination 23°

Ans. 6h. 5 1m. 408.

[178a] . In what latitude N. will the difference between

the longest and Shortest days be just 6 hours?

Ans. Lat. 41° 24’ N.

[179a] . At a certain place, when the sun’s declination was

10° N it rose an hour later than when it was 20° N. re

quired the latitude. Ans . Lat. 5 2° 27 N.

[180a] . In what latitude N. will the Shortest day be justone-third the longest (declination 23

°

Ans. 58° 27 N.

[l8la] . And in what latitude will the Shortest day bejust «gths the longest ? Ans. 41

° 24’ N.

[182a] . If a Ship sail from a certain place a miles due

146 TRIOONOMETRICAL SURVEYING.

east, then amiles due south, and then bmiles due west, andreach the same longitude required the latitude of the place

arrived at.EXAMPLE.

and b= 150’

. (Fig. Ans. 85° 0

'so

”.

Tofind the true bearing of a Terrestrial object.

[183a] . Given theapparent and true altitude Of aheavenlybody

,its declination , and observed distance from a terrestrial

Object to find the true bearing of the Object from a given

station . (Fig.

1 . When the object is in the horizon .

In latitude 7° 5 1 ’ N. the Observed altitude of the sun’

s

lower limb was 10° 30’west ofmeridian , and Observed dis

tance ofhis nearest limb from awell-defined point of land on

the same level with the eye and to the right Of the sun was

95 ° 16’

index correction Of the altitude sextant was 50"

and that of the other was 1 ’

the correction for heightOf the eye (14 feet) in taking the sun

’

s altitude was 3’

41”

required the true bearing of the point of land, the sun’s de

clination being 22°24

'

S . and semidiameter 1 5 ’

Ans. Bearing, N. 19° 0

’

30 W.

2 . When the Object is elevated above the levelOf the eye

it is necessary to observe its altitude.

In the preceding example, suppose the Object Observed to

be amountain, the altitude of whose summit is 10

°required

the true bearing Of that point. Ans. N. 17° W .

Measurement of an Arc of the Meridian.

The system of rules and operations by which the

relative position ofany number ofpoints in a tract Of countrymay be determin ed, so that it may be delineated on a plane

surface, is called Trigonometrical Surveying . When the ex

tent Of country is not great, the subject involves little dith

culty ; but when a kingdom such as Britain or France is to

148 TRIGONOMETRICAL SURVEYING.

respect to the meridian by compass,or more accurately by

astronomical Observations (see Problem 183) and then cal

culate the distances of all the stations from such meridian,and also the meridional distance Ofone station from the other.

down in the plan independently Of the others, and also the

direct distance between any two pointsmay be easily found

The manner of proceeding will appear from the following

(Fig. Let A,B,C, D,

E, F, be six stations

connected by four trianglesABC, BCD,BDE

,EDF the ah

glesare :

BAC=79°20’ CBD=39°20’

ABC : 5 1 31 BCD=69 28

ACB=- 49 9 BDC=71 12

DBE=45 28 EDF= 62 3

BDE=72 3 DEF= 52 25

BED=62 29 DFB=65 32

A Side,AB , Ofone of the triangles is 4213yards, and itmakes

with the meridian, NS, an angle SAB=62°5 2

’

at the point

A ; aand the station F makes at the poin t S an angle ASF

with themeridian=— 5 2° It 18 required to find the points

in which perpendiculars from the stations will cut the me

ridian, and the length of each perpendicular.

SOLUTION.

1 . Draw Bb, Cc, Dd, Ee, Ff, perpendicular to the meri

dian, and Bn, Dq, parallel to it, forming the right-angledtriangles ABb, BCm, EDn, EDp ,

FDq. Because the angles

of the four trianglesABC ,BOD, BDE, DEF, are given, and

alsoAB,a side of one of them, the five lines AB

,BC

,BD

,

DE,DE,maybe found from AB and each other (their logar

ithmsare

AB=— 3°624591 , BD=3°7335 5 9, DF=3

°578533,

BC=— 3°738255 , andDE=3°

2 . In the right‘angled triangleABb, the SideAB (on its

TRIGONOMETRICAL SURvEYING. 149

"log.) and the angle BAb=62

°52

'are known , hence we find

Ab= l92 1 °

,4 Bb=3749 °3 yards.

3. If from ABm=1 17° 8' (the supplement of BAb) theangle ABC=5 1

°31

'

be taken, there remains CBm=65°

therefore, in the right-angled triangle CBm the anglesand thelog. of BC are now known, hence we find Bm=2259 °6, and

Cm=4985 °2 .

4. And if from CBm=65 ° 37’

CBD=39° 20’ be sub

tracted, there remainsDBn=26° therefore in the right

angled triangleDBn theangles and the log. ofBDare known ,and hence Buz z 4854°7 and Dn=2397°6.

5 . From BDq=l 53°43

'

(supplement of DBn) subtractBDE=72° and there remains EDp=

- 81° then In the

right-angled triangle EDp the angles and the log. ofDE are

known, and thence Dp z 630'7, Ep=4306°1 .

6. Subtracting EDF=62° 3

'from EDp==81

° there

remainsFDq=19° 37

'and hence in the right-angled triangle

DFq we find Dg=3569°l ,

To determine the stationswe have now

Ab Bb

Ac=Ab+Bm Cc =Cm— Bb= l235 °9

Ad=Ab+Bn 67 Dd=Bb — Dn=

Ae=Ad+Dp = 74068 Re=Dd +Ep= 5657°8

Ff =Dd +Fq=2623°8

By these numbers, the position Of each station may be laid

down with great accuracy in a plan, independently Of the

others also the distance between any two may be readily

found ; for example :

(As

Lastly. We know by Observation that the bearing of F

from S is N. 5 2° 40’E. hence in the right-angled'

triangle

Efswe find Sf==2001°2 yards ; adding this to the line Af,

we have the length Of the meridian line between the stations

A and S= 12346°4 yards.

150 TRIGONOMETRICAL SuRvErING.

The abovemethod Ofmeasuring the meridian line between

two distant stations furnishes, as we have said,the means Of

finding the approximate length Of a dogma, and thence, on

the supposition of the earth being an exact Sphere, the magnitude of the earth ; for we have only to determine by Observation the difference Of latitude between the stations A

and S, and then divide thewhole length of the arc, found as

above, by the number of degrees contained in it the resultmultiplied by 360will give the circumference Of the earth.

Thus, let us suppose the latitude of A has been found bysome of the preceding problems to be 49

° 48’

6 4” N. and

that of S 49° 42’ N and the length of the line AS to

be 12346 4 yards ; then by a common proportion we easilyfind that the length of ad egree is about 69 miles. Picard,in 1670

, by Observations and measurements conducted in a

manner similar to the above, found the length Of a degree of

the meridian in latitude 49° N. to be 121627 yards, whichdiffers only 35 yards fromwhat is now considered as the moreexact length ; an accuracy

,however, which must be supposed

to be quite accidental.If we assume the above measurement to be correct, and

the earth to be a sphere, its circumference would be aboutmiles. Similar Observations and measurements have,

however, Since been made, in the same and other parts Of the

earth, with more accurate instruments fromwhich it appears

that the lengths of a degree of latitude are not invariable, as

would be the case if the earth were a sphere but that they

increasefrom the equator to the poles suggesting to uS the

figure Of an oblate spheroid .

The Several lengths of a degree of the meridian,measured

in Sweden,Russia

,England, India, &c. are given in the

Author’

swork on Navigation , Part II., and also the investi

gation of the following formula; for determining by means of

these measurements the lengths of the equatorial and polar

diameters Of the earth. By applying the formulae to the Oh

served lengths Of a degree in different parts Of the earth, we

APPENDIX

BEING

THE SOLUTIONS OF 180PROBLEMS

ASTRONOMY, SURVEYING,

NAVIGATION,

CONTAINED IN TRIGONOMETRY, PART I.

W JEANS,

LONDON

LONGMANS, GREEN,READER,

AND DYER.

1 878.

APPENDIX .

PROB. 1 (Fig.

THIS problem is Similar to the one solved in p.Let Cbrepresent the height Of the tower=2l6, Aa height

Of Observer’s eye= 5 , and ab the width Of rIver required.

Draw AB parallel to horizontal line ab, Bb=Aa=5 ,and CB=Cb— Bb=2l6 join AC ; then CABangle Of elevation : 47

,In the right-angled triangle

ABC are given the perpendicular OB=2I1, and angle CAB— 47° to find the base AB=ab, the width of river re

quired.

(Rule V. p . 57) cot.CAB,orAB=BC cot. CAB

PROB. 2 (Fig.

This problem is Similar to the one solved in p. 83.

In ABD,is

?) cot. B, or BD=AD cot. B=133°

7 (p.

In ADC,I

A?) cot. C, orDC=AD cot. C=71 °3

and BO=BB +DC=133°7 theAnswer.

PROB. 3 (Fig.

This problem is Similar to the one solved in p. 83.

To find BC (p. “Li—g tan. A, orBC=AB tan.A= 471 °1

ég=sec. A, orAC=AB sec. A=5 1 1 °9

The referencesare to Trigonometry, Part I., Sixth edition.

158 SOLUTIONS OF PROBLEMS

perimeter of feet= °224

of amile. 4 : 2 24 60°

: time required=3m Ans.

PROB. 4 (Fig.

This problem is similar to the one solved in p. 83.

In triangle BAC, AC=AR,andAD, the perpendicular

upon BC, bisects the angle BAC and Side BC .

DC=50, and angleDAC=75°are given, to findAC.

cosec. DAC, orAG=DO cosec. DAC=5 1 °77. Ans.

PROB. 5 (Fig.

Let AC represent the broken piece=39, A the pointwhere it struck the ground, and AB the distance Of its topfrom the base= 1 5 . Representing the known quantitiesACby a, andAB by b, and the unknown part BC left standingby :c, we have in right-angled triangle ABC, a

z=ag— b9=

(a— b)=54x 24 .

o

. x x (p. and

whole height of pole: 39 36=75 feet. Ans.

Pm . 6 (Fig.

Let D and E be the windows, DC=EC the ladder=36

a, BD height of lower andAE heightof other Draw FD parallel toAB, thenEDE is the angle of elevation Of E above D .

Let BC=x, AC=y, then x+y=breadth Of street.

InACE, CBD, y9=a9 (a— c) andaz

2=ai - b2

(a+b) . (a— b), whence y= 18°8,and a=30°7, as in last

breadth of feet.

2l

PROB. 7 (Fig. 17

Similar to problem p. 83.

160 SOLUT IONS or PROBLEMS

PROB. 12 (part OfFig.

Thisproblem is Similar to the one solved in p . 88.

Let AB represent the height of tower, D and C the stations ; then DO=100, ACB= 4G

°andADB=31°

are given, to findAB.

In Oblique-angled triangleADC are given DC=100,angle

'

ADB=31° and angle DAC=(ACB— ADC)=

14° to findAC.

(By p. AC DC : : sin . D sin. DACIn right-angled triangleABC aregivenAC : 202

, and

ACB=46° to find AB.

AB

RSin. AOB, orAB=AC Sin. AOB AB=145 °9. Ans.

PROB. 13 (see Ex. p.

The solution Of this problem is Similar to the last.Let AB (Fig 2 1 ) .

represent the height of cloud, D and Cthe two Ships ; then DC=880yards,ADC=35

°

,ACB=G4°

DCA= 1 16°, andDAC=29

°.

To find AC, or rather log. AC,AC=DC Sin. ADC Sin. DAC AG=3°O17503.

TO find AB,

ACSIn. AOB, orAB=AC Sin. AGB AE=935 °G

,

and height of eye above the sea=935°6 7 yards.

Ans.

PROB. 14 (part OfFig.

Let AB represent the tower, C and D the stations ; then

OB=2OO, DB= 350, and angle AOB=39°are given, to find

AB and angleADB .

ABTO findAB,

CB

TO findADB, tan.ADB whenceADB=24°

tan . ACB AB=162 feet nearly.

IN SURVEYING. 161

PROB. 15 (Fig .

LetB andC represent the two Ships,Athe Spectator ; thendistances AC=45 68 and AR=G852 (being 1 142 x elapsedseconds) ; hence in the triangle ABC are given two Sides, andincluded angleA=48

°to find the third Side CB.

Rule p. 54 or 5 5 the Side CB=5 147°9 feet.

PROB. 16 (Fig.

LetAB represent the Ship’

s mast, C the other ship’

s hull ;then, if BD is a horizontal line, DBC is the angle Of depres

sion,and is equal to the alternate angle BCA. In the right

angled triangle BAC are givenAB=8O, angle C=20°

; henceAC=219 °8 feet.

PROB. 17.

Draw two parallel lines AB, CD, perpendicular to a

horizontal line BD,to represent the two monuments ; let

AB=100, CD : 50, and through the tops A and C draw a

straight line cutting the horizontal line BD produced in E .

then angleE=37°

; also Since AB=2CD . BE - 2DE and

DE=BD. To findDE or BD. In the right-angled triangleCDE are given CD=50 and E= 37° ; hence DE and

BD=66°4 feet .

PROB . 18 (Fig.

Let AB represent the Obelisk, C and D the stations.

(1 ) In triangle ADC, given DC and all the angles, to

find AC.

In triangle ACE, given BC and AC and includedangle AGB,

to find AB,the height Of Obelisk=5 7

°62 feet.

PROB. 19 (Fig.

This useful problem is Similar to the one solved in p. 89 .

In triangleCAB are given theangles CAB,ABC, and

base AB, to find AC=706°8. In triangle DAB are


given theanglesDAB,DBA, andbaseAB,tofindAD: 13892 .

In triangle ADC are given the two SidesAC,AD, and

included angle CAD (the difl'

erence between CAB andDAB),to find CD= 1 174 °4 yards.

PROB. 20(Fig.

This problem is Similar to the preceding.

PROB. 2 1 .

Let ACB be the right- angled triangle resting upon itshypothenuse, C the right angle c= 100 feet. Let A be

the given angle: 36°

Let AC=x,and CD,

a perpen

dieulet from C upon base AB=y; andwhen the plane Of thetriangle is inclined at an angle Of 60

°

to the horizon, let theperpendicular distance of point C from the horizontal planebe denoted by s. In this case awill be the perpendicular ina right-angled triangle Of which y is the hypothenuse, andangle at base (By turning down the left - handupper corner of this leaf a right-angled triangle will beformed the creased line may represent the hypothenuse c

,

the corner Of the leaf the right angle C, and the other twosides asand y then , if this right-angled triangle be inclinedto the plane of the leaf at an angle=60

°

, the perpendiculardistance Of point C from the plane of the paper will be 2

,

the height Of the right angle from the ground to be calcu

lated .) In triangleACB, a=c cos. A In triangleACD,

y=az Sin . A Substituting in (2 ) the value Of a: in

we have y=c cos. A Sin . A Now the right-angled triangleAOB is inclined to the horizon at an angle=60

°

=y Sin . 60°= c COS. ASin . Asin. 60

° by substitutingthe value Of y from whence s= 41 °48. (See Ex. p.

PROB. 22 (Fig.

In triangleDBC, given DC= 100 D=33°

and

DBC=ACB — D=5 l°— 33°45

'=l7o 15 to find log. BC .

164 SOLUTIONS or PROBLEMS

BCH, given RH and CH, to find angle of elevation BCH17 47

PROB. 28 (Fig.

LetAand E be the Observers, C the balloon then angle

Of elevation CAD=20°, horizontal angle NAD=45°

(thebearing of C from A) , and horizontal angle AED= 1 1

° 15'

(the bearing of C from E). (l .) In horizontal triangleADEare given the Side AE=4000 yards, AED=1 1

°and

DAE= 180° to find log. AD. In vertical rightangled triangle ADC are given log. AD and CAD=20°, tofind CD, the height ofballoon .

PROB. 29 (Fig.

40°

and to find log. BC .

In right-angled triangle ABC, given log. BC andACB

to find height AB and distance AC of steeple.

PROB . 30(Fig.

In triangle BRS, given BR=54 feet, RBS=90°

— 4° 5 2'

—85°

and angle BSR=4° 52

'— 4° to find log. RS.

In right-angled triangle SAR, given log. RS, and ASR=NBS=4° to find AS the distance and height AR of

hill,and hence AB,

the height of top of building.

PROB . 31 (Fig.

Let the perpendicular CD be the breadth Of the river;then (1 ) in triangle ABC are given AB= 5OOand all theangles, to find log. BC. In right-angled triangle DBCare given log. BC and angle B= 79

°to find perpendicu

lar CD,the required breadth of river.

BC sin . AOram”Hi sin . C Sin . (A+R)

RG=AB sin . A cosec. (A+B) (1)and DO=BO Sin. B ; substituting the value of BC fromwe have DC=AB Sin. A cosec. (A+B) Sin . B.

IN SURVEYING. 165

PROB. 32 (Fig.

Let AB be the sloping base=340 feet, AD,BF, hori

zontal lines in the vertical plane ACD,and CD the height Of

hill required then angle CBF=46° CAD=40° and

EBF=EAD= 10°

CBE=CBF— EBF= 36° 10’

,andCAE=CAD— EAD

30° ACB=G°

In triangle ACB, given AB and all the angles, tofind log. AC.

In right-angled triangle ACD, given log. AC and

CAD, to find CD= 1221 ; adding height Of eye 5 feet, theheight of bil1=1226 feet.

PROB. 33 (Fig.

Angles observed with a Theodolite, p . 81

LetAbe an Object which we may suppose on . the topOfa

hill, and B and C two stations on its sloping Side. Conceivea horizontal plane to pass through the lowest station C, andlet Aa Bb be perpendiculars on that plane, meeting it in a

and b; join the points A,B,C in the oblique plane and the

points a, b, C in the horizontal plane, and draw BD perpen

dieuler to Aa ; then AaC, BbC will be right angled triangles,right-angled at a and b, and BDab a parallelogram .

There are given, BC the distance between the stations220 yards, horizontal angle bCa=62

° '

33’

(subtended at

C by A and B), horizontal angle Cba=70°15

'

(subtended atB by A and C), and the vertical angles ACa, BCb, the ele

vations Of the object A and the upper station B, to find the

horizontal distances ba, Ca, and the heights Aa, AD OfA

above 0and B.

In right-angled triangle BbC, given BO= 22O andBCb=8° 32 ’ perpendicularBb=32 °6and base Cb= 21 76 .

In horizontal triangle abC, given Side Cb=2 17°6, the

angle abC=70° 1 5 ’and aCb=62° to find ab= 263°1

and aC= 279° l . In the right- angled triangle AaC,

given the Side aC=279 °1 and ACa=32° to find


Aa= l75 °7. And Since Da=Bb=32 °6 AD=Aa

Bb=l43°1 .

PROB . 34 (last

Angles Observed with a Sextant, p. 81 .

The same construction being made as in last problem

(1 ) In right - angled triangle BCb, given BC=220 yardsand BCb=8° to find and Cb= 217°6.

In triangle ABC,°

ven BO=22O, and the three angles,

AC= 2 °5 18244, andAB= 299°5 . In right-angled

triangle ACa are given log. AC and angle ACa=32° 12

’

and Ca= 279 °1 . In right angled triangleABD, AD=Aa— Bb=l75 °7 and AB= 299 '5

PROB. 35 (Fig.

In triangle CADare given CD C= 5 7°

, and

and AD=487°27. In tri

angle ABD are given AD=487'27 and the three angles ;

hence AB= 349 °52, and BD= 498°7.

PROB. 36 (Fig.

(l . ) In triangle AFC are given CA= 3000, C=80°

and OFA= 25° to find FA= 6959 . In triangle

ABE are given FA= 695 9, AB=2000, and included angleFAB=34° 10' hence FB= 5422, the distance Of fort from

PROB. 37 (Fig. see Ex. p. 91 .

Describe a circle passing through the three pointsA, B ,

D, and join AR,EB. Then (by Geometry) Since angles in

the same segment are equal,

and EBA=ADE=12° 15

(1 . In triangle AEB, the Side AR= 5 35 , and angle

EAB=1 5° and EBA=12° . AE=2 438.

In triangle ABC, the three Sides are given, to findCAB=35 ° and CAE (=CAB

l68 SOLUT IONS OF PROBLEMS

PROB. 40 (Fig.

In triangle ACB, given angle C the course of the ship4 points or the angle A=45

°

(since B bears south-east

fromA), and Side CB=14 °5, to find the sidesAC and AB .

PROB.‘

4 1 (Fig.

In triangle AGB, tlfe bearing Of B fromA is east, and Of

C fromAisN.N.E. . angle CAB=6 points=67°

alsoangle ABC=4 points, and side BO= 10; henceAB, the distance Of boat from Ship, can be found.

PROB. 42 (Fig.

In triangle ACB, the bearing of C from A isN.E. b. N. ,

and Of B fromA is E. b. N. . angle CAB (the differencebetween E . b . N. and NHE b. 4 points, orAgain, the bearing OfA from B 18 W. b. S

, and OfC from

B isW.N.W. angle ABC=3 points, and Side AB=15miles; hence the other Sidesmay be found.

PROB. 43.

In any triangleABC, let AB=50; make the angle A=7points, and angle B= l point then if B is due south ofA,

and C is to the westward OfAB, the bearing of C from A isW. b. S. and from B, N. b . W.

In the triangle ABC are given the Side AB and all the

angles ; hence the SidesAC and CB may be found.

PROB. 44 (Fig.

(l . ) In triangle CAB, given AB=23 miles, CAB= 10§points,and CBA=3§ points CB=53°01 . In triangleABH , given AB= 23 miles, ABH=8 points, and HAB=4points points

,and BH=AB=23 miles.

In triangle CBH, given CB,RH ,and included angle CBH=

4} points, to'

find OH=42 °33, the distance ofcape from headland and BCH= 2} points. Hence C bearsfromH 21 pointsto the left, i. s. 2} points left ofW.N.W. orWAS.

IN SURVEYING.

PROB. 45 .

This prpblem is Similar to the last, the position and bearing of the bodies only being

'

different. Make a figure to suit

the case (roughly or by scale) , by drawing from a point Athe lineAC the lineAB N.W . ,

andAH due north.

Let AB : 6, and through B draw BC W . b. S.,and EH N.E. ,

and join CH.

In CAB,find CB=4 °13. In ABH ,findBH=6.

Lastly, in triangle CEH are given two sides, and in

cluded angle OBH=I3 points, to find OH=9 °71 miles, and

angle CHB=$ 13° 40

'

Hence C bears from B 13° 40

’ to the right, and therefore bearing of C from H is S . W. , or

S . 58°40

'W.

PROB. 46 (Fig.

Let Aand C represent the stations on deck, H the Objecton Shore ; then angle A=3 points (the bearing OfH from C),and angle points (the difference of bearings at the twostations). In the triangle, all the angles are given, and alsothe Side AG= 100 hence the distance AH and CH may befound.

PROB. 47 (Fig.

Let A be the required position Of the Ship ; then in

triangle ABC, the bearing ofC fromB is E N.E.,andAfrom

B is due south, the angle CBA=10points. We have

given, BO= 96O, AB=2000, and included angle ABC=

10points, to find the angle BAC=20° hence the bear

ing Of church isN. 20°32

’E.

PROB. 48 (Fig.

Let Abe the required position OfaShip then in triangleACH ,bearing Of C from H isW. }S. and OfA from H is S.

b. EHQE angle CHA=9§ points. In triangleACH we

have given, the two Sides GH=4 °23, and AH=2°3, and in

cluded angle CHA, to find angle HAC=4§ points ; hence


the bearing of C 18 4} points to the left of H, that is, to theleft OfN.b. and the requiredbearing is .

°

PROB. 49 (Fig.

(l .) In horizontal triangle ADS, given AS= 2640, SD800

, andangleASD=3 points=33° to findAD= 2024°2 .

In right-angled triangle ABD,AD and ADB

— 5° to find BA the height= l92 yards.

PROB. 50 (Fig.

In triangle ABC, the angle ACB (the difference of beer

ings OfAand B from C) is 3 points, the bearing OfA fromB is due west, and of C from B is S.w.b.W angle

ABC=3 points, and side AC= 10miles hence BC may befound.

PROB. 5 1 (Fig.

FromD, the highest part Ofmuzzle, drawDG parallel to

axis AE Of piece, then angle BDG=BCA, the angle re

quired.Through B draw a line perpendicular to axis, cutting

DG in H, andAB in K ; then in right-angled triangle BDHare given BD=

— 772 5 inches, and BH (=EX— DE)=62 2— 4 °92 : 1 °3 inches, to find angle EDG=BOA,

the angle te

PROB . 5 2 (Fig.

ED 50In triangleABD cot. AED

AD 175

AED=74° 3'

and AEB=148° 6’

30"=angle Of the

In triangle BAK aregivenAB=35O, BK=100, andincluded angle ABK(=90

°— 74° 3'

56’

to

findAH=25 5 °33=AG, the line Of defence.

Since the angle Of aregular

+ 2 BAH or 5 HAH ,=54°— 15°

56'

3'15

”HAH

,=76° 6’

30 the flanked angle.


Let AB=23, CD=y then AC=z + 5 and EF=y+ 5 ,and z +y+ 5= height ofmaypole required.

In FAE and ACD, and

202=y2 By (1 ) By (2 ) x2+

10x+ 425= y2. Subtracting 1) from (2) we get x+y=45

the height.

Pm . 5 5 a (Fig.

Let BD the horizontal distance OfAfrom B=x,andCD

the horizontal distance Of A from C=y, AD the height ofthe bill=z.

Let elevation of A at and horizontal angleBCD (the bearing ofA from and 0==ACD,

the elevation ofA from C required.

In right-angled triangleBCD, x=y Sin . B In verticaltriangle ADB, z

= x tan. a. In vertical triangle ADC,z=y tan . 6 (3) y tan. tan . a=y sin . 13 tan . a

tan . 0= sin. 3 . tan. a, and

PROB . 5 6 a (Fig.

(l In the triangleABC, given the three Sides, tofindone

Of the angles at the base, as B ; and thence in the right-angledtriangle ADB,

the perpendicularAD is easily found.

Or thusLet AD=x ; and ED=3; a, b, c the three sides of tri

angle then a= 4o,b=30, and In right-angled tri

angle ABD,-f- y

2=oz In right-angled triangle ADC,afl+ (a—

y)2=b2 Subtracting (1) from (a —

y2

— bg— 02

1 = l3°75 , and the per

pendicular required.

PROB. 57 (Fig.

(l .) In ABC are given all the angles and Side AB, to

find AC .

In ADC are given all the angles and side AC, to

find DC .

IN SURVEYING. 173

Or thusLet DAC= aa 23° 50

'

CAB=B=93°4

’20

ABC=7= 54°28

’Side AB=ac= 416 ; AC=x ; and

Bc=g.

2 Sin . 7 sin . 7(l . ) InABC,

a Sin . AOB Sin .

In ADC,

y=z tan . a y=a Sin . 7 cosec. tan . a 2787 .

PROB . 58 (Fig.

Let dist. BD of first station=x ; dist. CD Of second

station=y ; height AD of tower=z ; ABD=a=23°

and sideBC= 300. In ABD,x=x tan. a ;

a: tan . ,Bin ACD

,z=y tan . fl, “ x tan .

_atam a

’

a: tan . Band In BCD, §

= cos. EDO, cos. BBCtan . a

tan . 6 .

cot. a,and BDC= 25° 20’

45 Then in BBC,BD=

BC cot. BDC=633'4,and AD= BD tan . a=272 °7.

PROB. 5 9 .

In fig. 20,letAC represent the pole, inclined at an angle

A=SO°,B the place of spectator ; then AB= 100feet, and

angle B=54°

. In ABC are given side AB and all the angles,to find AC

,the length Of pole.

PROB . 60a (Fig.

Let AE (height Ofman) a=6 ; AB (height of column)a+ b= 200 BC (height of Let 0=ADE

or BDC, the angle subtended at D, the place of spectator, bya or0. Let x=BDE,

and y=-

,AD the width ofriverrequired.

aIn triangles AED,

ABD,and AOB, we have tan. 0

,

3/

a+ b+ c

tan .

3]tan.


tan. 0 y+y (2a+b) y

l — tan. (0+az) tan 6

a+ b+ c 2a 6_

y2(a:b) a substituting the values of a, b,

we have y=282°5 feet, the width Of river.

PROB. 61 a (Fig. to Prob.

Let AB= 23, AC=y, AD=z.

In CAB, y zas sin.ABC sin. ACB

InABD, z a: Sin. ABD sin. ADB

z=x sin. ABD cosec. ADB—1 1 1 85 23 .

In CAD,let CD=a=65 94

,let b= l '1852 and c

°8465 then z==bx, and y= cx,andangle CAD=85

° 46’=A,

are given, to find'

a: orAB.

In ACD,a M where tan. 0

“4” hav. A

8111 . 6 z—y

(Trig. Part II.) substituting the values ca: and bx fory ands,

we have tan. O=Mbx— ca:

a sin. 6

sin . 0 be hav. A

whence x orAB is found.This useful problem may be solved without haversines

thus In triangle

bz + ow b+c

bx— cx b— c

may be found, andthence the distance x,or AB=4694

yards.


EC . BACBAC

25 x 25 x8055 5

whence BAC= 53° 38’

and w= 100x tan . BAC=407°5 .

PROB. 63.

Since the area Of a parallelogram is equal to the productOf its base by its altitude,

’and the area of triangle=§ area

of a parallelogram with same base and altitude, area tri

angle=§ base x x 40x square yards.

Proof that area AD=AB AC.

(L) Let each of the sides

AB, AC of the rectangular

parallelogram AD contain

AH orAE, the linear unit,

an exact number Of times ;that is, let AB and AC be

com wnmrabk with AH .

Let the linear unit AH orAE=zv, and the squareAG=superficial

LetAB contain a linearunits=amAC

Then itmay be shewn that the mm ornumber ofsuperficial unitsin AD=ab, orareaAD=ab timesAG.

ADmultiplying AT }

ab

orADz ab timesAG.

If the Sides AB,AC be incommm wmble with AH, another

unit may be found which is commensurable with certain lines that

approach as near asweplease toAB a/nd AC, and therefore the product of the units in such lines will represent the area Of a rectangledifiering from the area AD by less than any assignable quantitythat is, we may in this case also express the area OfADby the product of the numberof units in AB and AC.

IN SURVEYING. 177

PROB. 64 a.

Let 23:— base Of triangle cut off,y=perpendicular; then bynote to last problem my=48; and as: y : 16 . . 4x=3y

and y=8 and

PROB. 65 a.

Let z=side Of equilateral triangle AB C . Then (Trig.

Part II. p. 49) area=¢8 s— a s— b s— c,where s

sincea=b=c=x area

we,or 180 z=20389

PROB. 66 a.

In any triangle ABC (fig. given base AB and angles

A and B, to find area. Find one Of the sidesAC, and

then,by Formula I. Trig. Part II. p. 49 , the area. Or thus :

AC sin. BIn triangle ABC, AB 8111 C

=m AC=T -AB

sin. B cosec. Now area=l. AB . AC sin . A

g-AB2 sin. A sin . B cosec.

PROB . 67 a (Fig.

In triangleABC, let a= 5 , b=3, and x=third side ; then

area2=s s— a s— b s— z=36, where

(Trig. Part II. p. (8

2x

) .

821

- 3:

whence a: (the

third side)=4 or J 5 2 by solving quadratic.

PROB . 68 a (Fig. 5 7

By the question, 2 .AEF=ABC. Let AD, alt. Of tri

angle=a= 10, base BC=x,EF=y, the base of triangleAFF,


and z= its an ; thenareaAEF=iABC3

g 3,

iax fl ). But by similar triangles ; Z.

2

:aj a 10

a 2—

T/-

2

whence distance from base= 10

ing the value of y in

PROB. 69 a.

Let z=the less side, andy=the greater side ofrectangle ;then 29: greater side of triangle, and 31 its hypothenuse

N/yfl— 4x2= third Side of triangle ; also 2x= side Ofsquare

whose area equals that of the two allotments 4m3=zy+

z a/3153— 4232 Length of paling for rectangle and tri

angle=2 (23+y) 2x+y+ N/yg — 4az

‘3,and length ofpaling for

square=8x 8x+ 5 5=2 (x+y)+ 2r+y+ N/y2 — 4x3

53:From equation (1) y 2

which substituted in (2) gives x : l 1

y=27°5 hence area and area tri

angle=a:~/y9

PROB. 70a.

Let a:+y, z , and x—y represent the three sides of the

right angled triangle then (cc—y) $2 and

—y)m= 2 16. From these equations the values Ofa: and y

are found, and thence the sides of triangle.

PROB. 71 a.

Let side of equilateral triangle=z area sin. 602

izz&J 3 (Trig. Part II. p. 49) :

:

Z-t- s/3 expense ofpaving

8z-

Z

z

s/ 3, and expense ofpalisading three sides

84 x 3x= 252x and


PROB. 75 (Fig.

LetCB represent length of ship= 160,Aplace ofObserverthen CAB= 20

’

also the angles SBC, SCE are nearlyequal to each other, since the Side CB (subtended by an

angle of a few minutes) is small compared with AB orAC.

either ofthe angles SBC or SCE maybe considered tobe the course= 22° 30’

(Nav. Part I. p.

AB Sin. C sin . 22° 30’

In mangle ABC’ B'

O'

sin . A sin . 20'

15'

AB=BC sin. 22°30

’

cosec. 20'

15 = 10247 feet= l °94

Otherwise (by circularmeasure).

Problems involving small anglesmay sometimes be solvedmore simply, by employing the circularmeasure Of the angle

(Trig. Part II. p .

Let AD: AB,join DB, then (the angle Abeing small)

each Ofthe anglesABD,ADB isnearly equal to aright angle

and DB may be considered as the arc subtending the angleA.

arc DB 20'15

Now circularmeasure Ofangle A=O fad—

AB—

7o .295 77

O .

AB=DEw but DB=BO sin . BCD (consider.

ing CDB as a right-angled triangle)=160 sin. 22°

ABl 60 sin. 22

°30

'

X X

12 15 x 1760x 3

to seconds, and dividing by 1760x 3 to Obtain distance inmiles)= 1

'94 miles.

reducing

PROB. 76 a (Fig.

Let AB represent the meridian , angle A=z°

,angle B

z°+ then C is the point where the shipsmeet

,and the

perpendicularCD= 100miles (the distance from the meridian

AB). The sidesAC and BC are to each other as 3 2 . Let

Ac=3y BC=2y.

3 2 taThen (Trig.

'

Part I. p. 5 3)31+ y n

IN SURVEYING . 181

tan . tan. 5 37’45

and difference Of lat. AB : AD+DE= CD cot,x+CD cot.

(mo+

or, 5

PROB. 77 a (Fig.

Let AB represent the height= 2 yards, AA’the diameter

of the earth ; draw the tangent BD then BDwill be the

distance seen by spectator (neglecting the effects ofrefraction,

By Geometry, BD2=BA BA’=2 x 14080000(if we

suppose diameter=8000miles= 14080000yards)BD=5307 yards.

PROB. 78 a (Fig. to Prob.

Let EH be drawn perpendicular to BA, the height ofmountain ; then E BD (the angle-Of depression of D, a pointin the horizon)=90

°- CBD=DCB, since angle D= 9O

°

DCB= 2° 13’Let a=BA=3 miles ; x=CA or

CB x+a a

CD, the radius Of the earth ; then see. CCl)

:

1:

1 +2

2 sin g-9

O 12

sec.

cos. C

2 x=a cos. C cosec 2 w,or diameter=7952 miles.

PROB. 79 a (Fig. to Prob. 77)Let DC orAC, the radius of the earth=a=4000miles ;

BA,height Ofmountain=:c. Then

(v

ia

—sec. C=see. 1

°

2 sin 2

sec. C — lcos. C

Cx= 2 a sin .

2

§cos. C=1 °402 miles.

SOLUTIONS OF PROBLEMS

PROB . 80a (Fig.

First solution (geometrically).

Upon the line AB=4OO yards, the distance between theobjectsA and B

,describe a segment of a circle containing an

angle equal to the given angle (35°

(Euclid III. and

measure a perpendicular distance CD= 5 60yards, the widthof the river ; through D draw EDF parallel to base AB, cutting the circle in E and F , then either of the points E, F

will correspond to the station on the other side Of the riverand the lines EA,

EB, or FA, FB,

measured on the same

Scale asAB,will be the required distances.

Second solution (trigonometrically).

(l . ) In right-angled triangle AGO (G being the center ofthe circle),givenAC= 200,and angle AGC=AEB=35

°

to find radius AG=B47°75 , and GC= 283°87, and angle

CAG=54°

GD=CD— GO=5GO

In right-angled triangle EDG are given GD=276'13,

radiusEG= 347°75 , to find angle EGD=37° 19

’30

Then the angle AGE= 180°

19’ — 107

°30

’

(5 . Draw GH perpendiculartoAE. We have now givenAG and angle AGB

— 5 3° 45'15

"in right-angled triangle

AGH , to fin d AB=%AE, ,whence AE= 5 60°1 .

To find the otherside, EB,there are given AE and

AB, and included angle 50

’

36° 14’

4'

to find EB=694 °37.

solution (analytically) .

By the geometrical construction Of the fig. (see first solution) we find that the two cmects are on the same side of thestation . Let A be the nearest object to station, B the moredistant.AB (distance of Objects) : =400; produce BA to K,

meetingperpendicularfrom station E,and letAK=a:,AE=y

BE= z,angle AEK=9, angle AEB between objects= a

35° and perpendicularEZK=5=560.


road feet) then plane Of triangleABDisperpendicularto road, and therefore ADC is a right angle.

In triangle ABC, AB=AC sin . ACB=AC sin. 20°

In triangle ADC ,AD=AO sin. ACD=AC Sin . 30

°

AB sin . 20° AB

”AD Sin . 30

° butLEI

—

ii0

sin . orADB=43° 9' 15

then in triangle ABD,AB=DB tan . ADB

— 200 tan. 43°9'15 feet.

Or thus (by Spherical Trigonometry).

The angle at C is a solid angle, formed by the vertical angleACB,

the oblique angle ACD, and horizontal angle BCDwhich call 0; let ACB=a, ACD=B. To findthe horizontalangle 0we may consider the point C the center of a sphere,and

'

the three angles, a,B, 0, may then be represented by thethree sides Ofaright-angled spherical triangle (Part II. p.

of which a and 0 contain the right angle, and B is thehypothenuse.

By Rule XIII. Part I. cos. B cos. a. cos. 0;

whence cos. a=cos. B see. a,

cos. 30°sec. 50

'

then in right-angled triangle BDC are given BD and angle

BCD,to find CB,

which with the angle ACB will give AB,

the height Of tower.

PROB. 82 a.

Let 7:c and 43: represent the sides, and included angle1 29

° 34'

let 0and go denote the other two angles.— l29°

(Ti-i

f. Part II. p.

tsn .

3m . 25

°13

'

which determines 6 e and therefore with 0 e alreadyknown, the angles 0and pmay be found.

IN SURVEYING. 185

PROB. 83 a.

First solution.

Let 0,20, 30denote the angles ; at , y, z the sides opposite

to them respectively ;then z +y+ z=6

sin. 0 sin . 30°

x

Z—

sin. 36 90°

— k Z

a: sin . 0 sin . 30°

5 l’

g} sin . 20 sin . 60°

—

W3

(3)Substituting in (l ) the values of y and z in (3) and

we have

whence and

Second Solution.

This problem,and problems 84

,85 , and 96, in which are

given (ormaybe found) the perimeter and all the angles, canbe solved in the following manner.

Let ABC be the triangle (fig.

To find side a. By (C), (D), Part . II. p. 49 ,

be sin.

2 12~(a+ b— c) t(a— b+ e)

“6 cos-2g §(a+e — b)

b b+ c)'

Ma+6 - c)

aebe cos}? g2 2 §(a+ c— b) “a+ b— c)

be sin }?1“a+ b-fl ?) §(a— b+ c)


PROB . 84 a.

Let 0, 2 9, 4 0denote the angles S the sidesoppositethem respectively ; then

42’

5 1”

22 sm 4 6 2 em 6 cos 2 92 cos.

3} sin. 2 6 sin. 2 0.

0

0o

z= 1 °247y.

J

O

B—

z sm

s

'

ig308° 6

—5 cos. 6= l '802

9Substituting in (l)we have1£02 +y+ l

°247y

y= 35°69 x= l9 °8, and z=44 °5 l .

Otherwise,see Prob. 83, second solution .

PROB. 85 a.

Let x, y, z denote the base, perpendicular, and hypothe

nuse respectively ; then :e+31+ z=24, x=z cos. 3 z,

y=z Sin . and z= 10°144

x=8'784,and y=5

°072 .

Otherwise, see Prob. 83, second solution .

PROB . 86 a (Fig.

Let the two sides AC and AB be denoted by a: and y ;then w+y=600, a=400, and A=SO

°

,are given to solve the

Atriangle. By (D) Part II. p . 49, we have my . cos.

2 —

2

} (x+y — a) my . COS.

2

100 xy==50000 sec.

2

Hence,knowing the values Ofx+y and my, the sides 3: and y

may be found.

PROB. 87 a.

By (C) Part II. p. 49,be Sin e -gt %(a+b— c) ya b+ e)

whered b=600,e= 400,andA=60°

sin }; Sin .

2 3o°=i

400 i — b+ 4oo) 400b=2oo(a— b)80000 2b=a— b+4o0, or3b— 400=a=600 —b 4b

1000 b=250,whence a=35o.


angles at the base, and a: and y the Sides Opposite to X and

XY. Then w+y= b, and X

— Y=d. Nowsin. is

, andsin . A a

sin. Y

sin . A a a

sm. X+ sin. Y b — Y) b

cos. b cos. id.

whence X+Y may be found, and with X— Y= d,the

values ofX and Y.

PROB. 93 a.

Let a be the base, and a: and y the other sides of thetriangle ; X,

Y the angles opposite to z and y respectively,

andA the angle opposite G.

Then x—y=d, and X— Y=D.

sin. X a: sin. Y_y Sin. X— sin . Y

sin. A a’ and

sin. A==a

subtracting,Sin . A

:c—y d 2 -Y) d

a a sin. (X+Y)2 cos. sin. £ (X — Y) d

a

Y)=gain . &D,and since X — Y is also

sin . fiX -i-Y)

known, the sides sud

d

angles may be found.

PROB. 94 a.

Let ABC (fig 20) be any triangle then angle A, side b,and the difference (a— c=d) between a and c are given, to

solve the triangle. By (C) and (D) Part II. p . 49,be smfl

é

2

“a+b— e) %(a— b+ c) and be . cos.

2A

2A (a+b— e) . (a+ o— b)

&(b+ e— a) tan.

(b+e— a)’ Similarly,

IN SURVEYING. 189

9 (a+ b — b)(b+ e— a

)2

2 — c)' .

tain fia+ b— c

C b+ e— a A b— (a— e)‘

E “ swims m ns . ;A.

PROB. 95 a.

LetABC (fig. 20) be any triangle ; then angle A, side b,and the sum of a and c are given, to solve the

C a+ o— b

C m— btan .

-

2 m+b°

PROB. 96 a.

See Problem 83 second solution.

PROB. 97 a.

In fig. 21 , produce BCD to a point E,'and join AE.

Let AB represent the object, and C, D, E the three sta

tions.

DC=a, ED=b, the angle ACB=20, ADC=90o— 0

, and

AED=0.The triangle AEC is manifestly isosceles, since angle

ACB=AEC+EAC, or 20= 0+EAC, EAC==0, or AO

EC= a+ b.

AE sin. 20In AEC,

a+b sin. 00

O

In AED,

AE_ em . (90 0) cos. 0AE

b — 20) cos. 20

sin. 0

190 SOLUTIONS os PROBLEMS

cos. 20 CB=§ b hence AB2=AC2 — CB9

(a+ b)z height (a+ b— 5b)

x/(a-f-gb)

If a=b=20, then the height : x

PROB. 93 a (Fig.

Let the angles BDC, ADC be denoted by a, and Bre

spectively, and the sides Of the triangleABC by a, b, c, in theusual manner. Let angle CAD=x

,and GED then in

triangles BDC and ADC, we have

b sin . :e sin . a: a sin . Bsin . B

"

sin . g b sin . a

sin . x+ sin . y_ a sin . B+b Sin . a

sin . x— em . y a sin. B— b sin . a

b sin . a

a sin . 66 sin . a

a sin . B

whence tan .Ma:— y) tan . b(x y)

and ze+y, the'Sum Of the angles, is known, since it is=360

°

C -a and by the equation . x—y is given, hence the

angles a: and 3] may be found.

Applying this expression to an example, suppose thata= 5 4° 13

’

B= 33°

7 42'

and that log. a.

log. b= 4 °02 1 189.

First find the value ofb sin' a

,which is 10329

,a sm. B

then tan . fi re—y) tan.

- 329or tan .

JAE —

y) tan . 97° 5

’

5 520329

(placing the proper Sign Over tan. 97° 5

’

and thusdeter


PROB. 100a.

describe squares, the centers of which are A’

, B’

,C ’

; jointhe pointsA’

, B’

, C'

,forming the equilateral triangle

A’ being Opposite'

A, &c. , then area’

ABC

(1 +1J 3).

For areaABC=}ACQ. sin. 60°

1 12

area areaABC .

Now to find the value of B ’

C'in terms ofAC.

In triangle AC” B’C

’2

or-. 2AB"Z— B’

C’2

B’C

’9=2AB’2 (1 cos.

2AB ’ 9z(1 5

But AB'2= 2 . (A

2

O

) = ;Aoehence area +5s/3) areaABC.

PROB. 101 a.

The geometrical construction Ofthisproblemmaybemade

To find the course and distance of the vice-admiral.Draw N.S. andW.E. at right angles, and letA, the point of

intersection, be the place Of the admiral. Make the angleNAV=2 points, or and take from a scale AV : G;

then V is the place of the Vice-admiral. Make SAB=4points, or S.E., andAB=3, then itwill simplify the problemifwe suppose the admiral to be transferred to B, and to sail

due east along the line BV’

,instead of alongAE, which is

the actual line described by the admiral ; for then the problemwill be reduced to this to find the course and distanceof the Vice-admiral, so as to formajunction with theadmiral.Join BV, and take BC=G, the rate Ofadmiral

’ssteaming, and

with C as a center and radius= l0 (the rate of vice-admiral)describe an arc Of a circle cutting BV (orBV produced) in

IN SURVEYING . 193


D ; join DC . Through V draw VV’ parallel to DC ; then

W ’is the Vice-admiral’s distance, and the angleANY which

it makes with the meridian N.S. is the course .

rate of admiral BC BV’

V’is the point of junction , on the hypothesis that the

admiral sailed from B . But the admiral steamed along the

line AE, and described AA

’=BY ’in the same time ; A’

is the position Of the admiral when the Vice-admiral is at V’

,

and it is evident the position ofV’

from A’

is the one re

quired, namely, S E . 3 miles.

TO find the course and distance of the look-out steamer.

Make the angle SAL : 3 points or and AL= 10miles

, then L is the position Of the look-out with respect tothe admiral. Take AA”= 5 miles (the distance the steameris to be a-head Of the admiral), and suppose, as before, theadmiral to be transferred toA

”

; join LA” then the problem

may be solved by supposing the steamer to form a junctionwith the admiral, asin the former case.

From A”E take A”

G’=Gmiles, the rate of admiral, and

with C'

as center, and distance 14 miles, describe an are cut

ting A”D’

in D ’

, and join D’C’

. Through L, the place Ofthesteamer

,draw LL

'

parallel toD’C’

. Then LL’

is the steamer’

s

distance run to get into her station,and the angle it makes

with the meridian is her course.

rate Of admiral A C A”

L’

rate of steamer C ’D ’

L’

L

L'

is the point of junction Of the admiral and steamer ;but since we have supposed the admiral transferred toapointA”

, when in fact he was atA, the position Ofadmiral will be5 miles to the west of the steamer.

TO find the course and distance Of reinforcement. JoinAV ’ then AV’

is the distance run, and the angle V’

AS the

course of reinforcement.

This problem may be solved trigonometrically, thusTo find the course and distance of Vice-admiral.

196 SOLUTIONS or PROBLEMS IN SURVEYING.

TO find the course and distance run Of the reinforcement.TO find BV

’orAA’

thedistance run by admiral.BV

’6 10 BV

’=5 °62

Or thus : in triangle VBV’

are given all the angles and

side VV’

to find BV’.

TO find distance AV’

.

In triangle ABV’are given two sides, AB and BV

’

, and

included angle ABV’

, to findAV’=8 miles.

TO find course of reinforcement.In same triangle find angle BAV

’= 29° 44'

left of B.

bearing ofB, S.E.=45 0 left Of S.

bearing Of reinforcement=74 44 left Of S.

hence course Of nearly.

TO find time Of reaching their appointed stations.

10 l h. m= 5 6 min. for vice-admiral14 1 h. E= 38 min . for look-out.

To find the rate Of steaming Of reinforcement, we have5 6m. 1 h. 8 z=8§miles.

SECTION II.

PROBLEMS IN ASTRONOMY,&c.

PROB. 102 (Fig.

LETNWSE represent the horizon,NZS the celestialmeridian ,WZE the prime vertical (see definitions,Trig. Part I. p.

take NE=5O° the latitude of spectator ; then P is thepole Of the heavens, since the altitude of the pole=lat. Of

place (p. and PZ is the colatitude ; take PQ=90°

,

then Q is a point in the celestial equator : throughEandW,

the east and west points, and Q draw the circleWQE ; thiswill represent the celestial equator (p. l Let X representthe place Of the sun at the time Of Observation, and throughX draw XZ a circle Of altitude, andiXP a circle of declination : then in the spherical triangle PXZ aregiven, the polardistance PX the zenith distance XZ— altitude), and the colatitude Of spectator PZ — lat ),to find the angle PZX, the azimuth Of the sun (p ;Then by Rule VIII . p. 62 , PZX=1 1 1

° 5 1’

(see Ex. p.

Hence the azimuth is N. 1 1 1° 5 1

’ W. , the angle being measured from the north towards the west.

PROB. 103 (Fig. 7

The same construction being made as in the precedingproblem, there are given the three sides Of the triangle PZX,

to find the hour-angle Of the sun ZPX, which is also theapparent time, since the heavenly body is west Ofmeridian


(p. Whence (RuleVIII. ) ZPX==2h5 7 the hour

angle required.

PROB. 104 (Fig. 7

In the fig. are given the polar dist. PX, the zenith distance ZX, and the azimuth PZX,

to find the hour-angle P ;this is easily done by Rule X . p. 68. Or thus

sin . P sin . zen . dist. cos. alt.

sin. az. sm. pol. dist. cos. decl.

Sin. F=sin . az. cos. alt. sec. decl.

Whence P=2" 16'

(see Ex. p.

PROB . 105 (Fig. 7

In the fig. are given, the polar distance PX=74°

cc

latitude hour-angle P=38°

to find the zenithdistance ZX,

and thence the altitude.

Calculation by Rule IX. p. 65 ; see Ex. p. 131 .

PROB . 106.

Construct a similar figure to the one in the precedingproblem,

but place the heavenly body on the east side Of themeridian, since the hour-angle is A.M.

In the fig. are given, the polar dist . PX : 77° colatitude PZ= 39° hour-angle P : 2h. 53m. ls. ; to find the

angle PZX, the azimuth . Calculation by Rule XI. p . 69 .

In the following twelve problems the heavenly body issupposed to be on the meridian at the time Of Observation ;in this position the triangle PZX (fig. 73) vanishes, and thecalculations are Simplified.

PROB. 107.

Describe the horizon NWSE,themeridian NZS, and the

prime verticalWZE take Sm=7 to represent the sun’s


PROB . 109 (Fig.

Describe as before the horizon , celestial meridian, andprime vertical.

,Take a distance Nm on the meridian= 7

to represent the star’

smeridian altitude then m is the place

Of the star. (The altitude in this case ismeasured from N,

the north point Of the horizon, since the zenith Of the spectator is by the problem south of the body. ) And Since the

declination is 25° north, take ,mQ=25

°

, measured from msouthward, and draw WQE, the celestial equator.

Lat.=ZQ=mQ— mZ= star’

S decl. - zen. dist.25

°

The latitude is north, since by the fig. the zenith is northOf the equator.

PROB. 1 10 (Fig.

Describe horizon, celestial meridian, and prime vertical.Take Nm=30°, the sun

’smeridian altitude (measured from

N, since the zenith is south) take mQ= the sun

’s

decl. , then a circle through WQE will represent the celestialequator.

Lat .=ZQ Zm mQ zen . dist. decl.— 60

°

The latitude is south, since the zenith is, by the figure,

south Of the equator,and therefore the spectator is in the

southern hemisphere.

PROB. 1 1 1 (Fig. to Prob.

Describe the horizon, celestialmeridian, and prime vertical. Since the latitude is 40° N. take ZO= 4O

°

, and drawWQE to represent the celestial equator : from Q measure

Om=20°

, then m is the place of the heavenly body,since

its declination is 20° N and

mZ=zen. dist.=ZQ Qm=40°

altitude Sm=7

IN NAUTICAL ASTRONOMY. 201

PROB. 1 1 2 (Fig.

Describe horizon , celestial meridian, and prime vertical.

Take Nm: then m is the place of the sun (the altitudebeing measured from the north, since the zenith is south ofthe heavenly body) take ZQ= 50

°=the latitude of place,and through Q draw WQE to represent the celestial equator

(ZQ is taken to the north Of Z,since the place is in the

southern hemisphere) : then mQ is the declination Of body ;and

mQ=mZ — ZQ zen . dist . — lat=60°

The declination is 10°north, Since by the figure the

heavenly body is north of the equator.

PROB. 1 13.

Describe the following parts Of fig. 66, the horizonNWSE, celestial meridian NZS, and prime vertical WZEthen since the place is on the equator, the celestial equatormust pass through the zenith

,and therefore WZE represents

also the celestial equator : take a distance Sm= 57°, to repre

sent the star’

s meridian altitude (measured from S,Since the

zenith must be north of the star) then

mZ=33°= zenith dist.=declination,since the equator passes through Z.

The declination is south, m being south of Z.

PROB. 1 14 (Fig.

Let m be the place Of the heavenly body belowpole P

,and m

lits place above pole.

Then Nm= star’

s meridian altitude at inferior transit.Nm

lsuperior transit.

and mP=mlP= star

’

s polar distance.

Nowlat.=NP=Nm

1— m

lP

2 lat.=Nm+Nm1 (since mP=m,P)


PROB. 1 15 (Fig.

Describe horizon and celestial meridian. Take Nm20°=star’

smeridian altitude at inferior transit and since It

passes through the zenith at superior transit, bisect Zm in Pthen P is the pole, and (as in last problem)

lat. PN N

PROB. l 16.

Construct the horizon NWSE,and celestial meridian

NZS. From N take a meridian altitude=x, and from Z a

meridi’an zenith distance a:

meridian altitude at superior transit= 90°— :e

lat below pole+alt. above pole)

PROB. 117.

Describe the horizon NWSE,and celestial meridian

NZS ; on the meridian, take NP= 60°

, the latitude ; thenP is the pole ; and since the declination is the polardistance= 50°; take therefore fromP towardsNadistance Pm

and from P towards the zenith a distance Pml= 50°;

then m is the place Of the body below the pole, and ml isthe place Of body above the pole.

Alt. below pole=Nm=NP — Pm=60°

alt. above — Zml=90° — (Pml — PZ)

90°

PROB. 1 18.

In fig. 70, let mZ represent the parallel Of declinationdescribed by the star. ThenPZ=39° 12 ’=star’

s polar distance . decL= 50°48

'

N.

and alt. below pole=Nm=NP— Pm=50° 48’— 39° 12

’

1 1°

PROB. 1 19.

Since the sun, in this problem,touches the horizon at

the inferior transit, its polar distance must be equal to the


and take a point V six hours from the meridian, and drawPV the pol. dist. , and ZV the zen . dist. then

sin. eli.= sin . deel sin lat.

Proof: In triangle PVZ , right -angled at P, mark zen .

dist. VZ , pol. dist . PV, and cO-lat. PZ then

cos. VZ= cos. PV . cos. PZ

or sin . alt.= sin . decl. sin. lat.

Again ,cos. azimuth=cot. lat. tan . alt . (z)

Proof. In a similar triangle, mark PZV the azimuth,cc-lat. PZ, and zen . dist . VZ then

cos. Z= tan . PZ cot . VZ

or cos. azimuth=cot. lat. tan . alt.

Again, cos. lat.= cot. decl. cot. azimuth (n)

Proof. in a similar triang le, mark co-lat. , pol. dist. , andazimuth ; then

sin. PZ : tan PV cot. Z

or cos. lat.= cot. decl. cot. azimuth.

Again ,cos. decl.=sin . azimuth cos. alt. (0)

Proof. In a Similar triangle, mark pol. dist., azimuth,and zen. dist. then

sin . PV= sin. Z sin. VZ

or cos. decl.= sin . azimuth cos. alt.

The heavenly body in the horizon, as at D ,fig.

p. 125 .

Draw the horizon and celestial meridian as before, andtake D a point on the horizon

,and draw PD the polar dis

tance, and ZD the zen . dist . then

cos. hour-angle- tan . lat. tan. decl. ( 7)

Proof. In quadrantal triangle PZD, mark hour-angle

P,cc-lat . PZ

,and pol. dist. PZ then (ZD being

cos. P — cot. PZ cot. PDor cos hour-angle

— tan . lat. tan. decl.

Again, sin . decl.= cos. lat em. amplitude.


Proof. In a similar triangle, mark PD,P2

,and z the

co-amplitude ; then

cos. PD : sin. PZ cos. Zor sin . decl.= cos. lat. sin. amplitude.

Again, sin. lat -cot. amplitude cot. hour-angle (A)Proof. In a Similar triangle, mark PZ, Z, and P ; then

— tan . Z cot. P— cot. amplitude cot. hour-angle.

Lastly,cos. amplitude=cos. decl SIn . hour-angle

Proof. In a similar triangle, mark Z, PD, and P

Sin . Z= sin. PD Sin . P

or cos. amplitude=cos. decl. Sin . hour-angle.

In the above formulae the lat. and decl. are supposed tobe Of the same name. When they are Of different names,the last four formulae (x), (A), may require somemodification : this may be easily done by remembering thatPD instead Of 90

°— decl.

,and angle Z=90

°+

amplitude instead Of 90°

ampL, as in former case. The

other formulae require no alteration , since they are appliedto problems in which the latitude and declination are alwaysof the same name.

PROB. 120.

Describe the following parts of fig. p. 125 , the horizon,celestial meridian, and prime vertical. Let Y be the placeof the body on the prime vertical, Z the zenith of spectator,and P the pole then in the spherical triangle PYZ, rightangled at Z, are given PY (the complement of the decl. )

and YZ (the complement of to find PZ

the complement of latitude, and thence the latitude.

By Rule XIII. p. 72,

cos. PY=cos. PZ cos. YZ (see Ex. p . 7

PZ=46° 50’

and lat.=43° 9’30

” N.


PROB. 121 .

Describe fig. to prob. 120. In this there are given, colat.PZ=90

°

50° or39°

pol. dist. PY=9O°

or 66° 32

’

15 to find zen. dist. YZ (and thence theand hour-angle P.

Mark PY and PZ as,known Mark PY and PZ as known

quantities, and YZ as un quantities, and P as un

known ; then, Rule XIII. known ; thencos. PY=cos. PZ cos. YZ cos. : cot. PY tan . PZ

YZ=59° and hour-angle=4h. 37m. 4S.

alt.=30° 5 5’

If the student is acquainted with analytical trigonometry

he can simplify the above and similar formulas that will frequently occur in the following pages, before he Opens the

table Of logarithms. Thus :

Since cos. PY=cos. PZ cos. YZ

sin . decl.=sin. lat. sin . alt.

Sin . alt.=sin. decl cosec. lat.

cos. P.= eot. PY . tan . PZ

or cos. hour-angle=tan . decl cot. lat.

From these two expressions the altitude and hour-angle

found directly from the quantities given in the question.

PROB . 122.

Make a right-angled triangle similar to PYZ in fig. p. 125 ,but so that Y may be to the east Ofmeridian, somewhere on

the line ZE ; let Y be the place of the star whose altitude

YE is take another point Y’ between Z and Y to re

present the place of the other star, whose altitude is

and join PY’ then in triangles PYZ, PY

’Z,are given,

colat. PZ=39° and zenith distances ZY and

ZY' to find the hour-angles at P, and thence YBY

’

their difference.


PZM= lOO°,and ZM : 90

°

,to find hour-angle P, and

thence apparent time (def. p.

By Rule XIV. p . 75 , cos. PZ — tan . PZM cot. P

placing the proper signs over known quantities, by Rule,p. 29, we find P is less than or 6 hours,

whence P is found= 5h. 28m. 5 38.

app. time= 18h. 31m. 7s. , or 6h. 31m. 7s. A.M.

Simplifying formula cot. P=siu . lat. cot. PZM—sin . 50

°cot. 80

°

PROB. 126.

In fig. p. 125 , suppose D the place of the sun on the

horizon ; then amplitude=WD=37° Complete quad

rantal triangle PDZ then in triangle PDZ are given, anglePZD= 5 2° 30’

(complement Of amplitude) , pol. dist. PD= 74° and ZD=90°, to find colat. PZ , and thence thelatitude.

By Rule XIV. cos. PD=sin. PZ cos. PZD

whence PZ may be found.

Or simplifying formula sin. decl.=cos. lat. . sin . ampl.

cos. lat.=sin . decl. cosec. ampl.

PROB . 127.

Construct the following parts offig. p. 125 , horizon, celestial meridian, prime vertical suppose D2 to be the placewhere the sun rose, and D the place of setting ; join PDand ZD then in quadrantal triangle PZD there are given,colat. PZ , pol. dist. PD, and ZD=90

°

,to find angle PZD

(the complement of amplitude), and hour-angle ZPD (thehalf the length of day), and thence the lengths of the dayand night.By Rule XIV . cos. PD=sin . PZ cos. PZD

cos. P — cot. PZ cot. PD

placing the proper signs over the terms (p. we shallfind cos. PZD is positive, and cos. P negative,


whence PZD=59° 5 5 ’

and ampl.=W . 30° 4

’

30 N .

P=7h. 36m . 4ls.=angle PZD

therefore the sun rises at 4h. 23m. 19s. A.M . , sets at 7h .

36m. 418. PM,length of day 1 5h. 13m. 22s.

,length of

night 8h . 46m. 38s. , and amplitude at rising and setting

30°4

’

30 N.

Formulae simplified sin . ampl. : sin . decl. sec. lat.

cos. hour-angle— tan . lat. tan . decl.

PROB. 128.

Since the day is more than 12 hours long, the sun mustrise (in north lat. ) to the north Of equator, as at S in fig. 97.

Construct fig. then in quadrantal triangle ZPS are given ,

colat. PZ=39° hour-angle P=7h. , and ZS= 90°, to

find PZS, the complement of amplitude.

Rule XIV. cos. PZ — cot. ZPS tan . PZS

whence PZS=N. 70°5 5

’

30”

E.

and ampl.=E. 19° 4’30

" N.

Simmified cot . ampl.— sin. lat. tan . hour-angle.

PROB. 129 .

Describe horizon, celestial meridian , prime vertical, andWQE,

the celestial equator,fig. 71 . Take some point X on

the equator to represent the sun’s place

,and draw circle of

altitude Z0,and circle of decl. PX ; then in quadrantal

triangle PXZ are given, zenith distance ZX=67° hourangle P=3h.,

and PX : to find colat. PZ , and thencethe latitude.

By Rule XIV. cos. ZX=sin . PZ cos. P

whence PZ=33° 26’

and lat.= 56° 33’30

Simplified cos. lat.=sin . alt. sec. hour-angle .

PROB. 130.

Let X,fig. 74, be the place of the heavenly body ; then

in triangle PXZ are given, zen . dist. ZX, pol. dist. PX, and


azimuth PZX, to find colat. PZ, and thence the latitude.

The quantities given are not sufficient to enable us to find

PZ by the common rules Of spherical trigonometry. In

cases of this kind it is usual to drop a perpendicular upon

one Of the Sides, or side produced, thus forming two right

angled triangles, and then to apply,if possible, Rule XIII.

Problems may Often be solved by means Of analytical trigo

nometry, when the common rules are insufficient, or tire

some if applied. We will give a few solutions Of this

problem, as they will be useful exercises in analytical trigonometry.

(1st solution.) By dropping a perpendicular, as XM, on

In right-angled triangle XZM are given XZ and

XZM, to find XM and ZM.

sin . XM= sin. ZX . sin . XZM XM=3S° 34’30

cos. XZM=cot. XZ tan . ZM ZM=30 1 2 45

In right-angled triangle XPM are given XP and

XM, to find PM.

cos. PX=cOS. PM cos. KM PM=61° 8’

45

whence PZ=PM— ZM=30° and lat.=5 9° 4'N.

(2d solution .) By means Of formula (U), Trig. Part II.

(L) Find angle P, by Rule X. Part I. p. 68, thus

sin. P : sin. PZX sin . XZ : sin . PX

P=42° 19’

30”

(2) By (U).

_ tan. %(PX — ZX) sin . 1}(PZX+P)W“ ’A’PZ "

sin . ; (s — P)

; BZ=I5° 27

'

and lat.=59° 4'

30 N.

(3d solution .) By means of a subsidiary angle.

Part II. p . 62 (L).

2 12 SOLUTIONS or PROBLEMS

cos. p cos. a;

cos. a1

cos. y= sin. decl. cosec. alt. cos. y

hence by (1) y: 30° 12

’45

” by (2) :e=6l° 8

’30

.e—

y the colat. : 30° 5 5

’and lat.=59° 4

’1 5

orcos. a= cos. p cos. y see. a

PROB. 131 (Fig.

In this important problem are given the polar dist. PX— 90

°— 22° the zen. dist. ZX= 90

°— 37° and hour

gle P= 2h. 15m.,to find the latitude=90° — PZ .

As in the last problem,the common rules of trigonometry

will not enable us to find the side PZ from the three knownparts Of the triangle. We may proceed as follows

By dropping a perpendicular XM.

(l .) In right-angled triangle PXM are given PX and P,to find PM. (Rule XIII. p .

cos. P= tan . PM cot. PX

PM= 63° 31 ’

Let h=hour-angle, d=declination , a=alt.simplified tan . PM : cos. h cot. d

In right-angled triangle PXM are given PX and P,to find MX.

sin . MX=sin . P. sin . PX. (Rule XIII.)

(or Sin. MX= sin . h cos. d) MX=30°

In right-angled triangle ZXM are given ZX and

KM, to find ZM.

cos. ZX=cOS. ZM cos. MX

(or cos. ZM=sin . a sec. MX) ZM=45° 2’15

PZ : PM— ZM : 18° and lat.= 7l° 31’N.

(2nd solution.) By means Of (U) , Trig , Part II. p . 62 .

(l Find angle PZX. sin . PZX : Sin P : sin . PX : sin . ZX

(or sin . PZX=cos. d sin. b see. a)PZX=40° 12 ’

or its supplement 139° 47’an

ambiguous result but it is evident by a'

rough constructionOf figure that the latter value must be taken . See fig. 74.


By formula (U), find §PZ (as in last problem)9°14

'whence lat.= 90° —PZ= 71° 31

'

(3d solution.) By means of a subsidiary angle.

Since (Part II. , p. 62 )

sin . PX sin . PZ cos. d cos. 0

sin. d sin. l+ cos. d cos. I cos. P= siu . a

or sin . d (sin. 1 cos. I cot. d cos. P)= Sin . a

Assume cot. 0=cot . d cos. Psin . d (sin . 1 cos. I cot. 0)=sin . a

orsin . d cosec. 0 cos. (l~0)= sin . a

cos. (l~ 0)= sin . a cosec. d sin . (2)Formula (1) determines 0, and thence, by l~0may

be found, and consequently the latitude I, as in the last

problem.

(4th solution .) By elimination .

Let PX=p, ZX=a1, P= h, PM=x,

ZM=y, XM= z ; then colat.=xl’

y

(1 To find a,or PM . In triangle XPM,

cos. h=cot. p tan . a: tan . a=cos. h tan. 1)

cos. b cot. doc] .

To find y, or ZM. In triangleKPM,

cos.p =cos. z cos. a. In triangle ZXM,cos. a

1cos. y cos. S

OS . cos. a:ehmmate cos. z by dIVISIon

c 19

cos. a1

cos. y

cos. y=cos. a1 cos. a see. p

sin. alt. cos. decl. cos. a: (2)Formulae (l) and (2) determine a and y, and thence the

latitude, Since colat. : :efiiy.

PROB. 132 (Fig.

Ex. 1 . Let X be the place Of the sun in the eclipticAXM

,A the first point of Aries, and ARQ the celestial

equator. Through X draw the circle of declination PXR ;then in the spherical triangle AXR,

right-angled at R, are

214 SOLUTIONS or PROBLEMS.

given angle A the obliquity of the ecliptic, and AR the

sun’s right ascension, to find AX the longitude, and XR the

TO find longitude AX. Rule XIII.

cos. A.=tan . AR cot. AX AX=64° 33' 15”

To find declination XR.

Sin. AR==tan . XR cot. A XR= 21° 4

’15 N.

Ex. 2 (fig. 77 In this example the sun’

s right ascension

exceeds 12 hours, it must therefore have passed the autum

nal equinox, or first point of Libra, to the south side of the

To find the place Of the sun, letALA1 represent thewhole

of the celestial equator, A and AIthe same point, namely,

the first point of Aries, ACLC,Al the ecliptic, cutting the

equator in the two poin tsA (or A1 ) and L. Take AR equal

to the sun’s right ascension, and through R draw the circle Of

decl. RX then X is the place of the sun , and its longitudeisACLX, and decl. RX.

To find longitudeACX,and decl. RX.

Subtract 12 hours from the right ascension ALR,the

remainder is LR ; and the angle BLK is also known, beingthe obliquity ; then in triangle BLK,

sin . RL= cot. Obliq. tan. decl. decl.= 22° 58’ S.

cos. Obliq.=tan. LR cot. LX LX= 78° 30’15

”

long. ACLX= 180°+ 78°30

'15 : 258

° 30’

15

PROB. 133.

Ex. 1 (fig. Let X be the place of the heavenly body,AR the celestial equator, and AM the ecliptic.

Through Xdraw PXR, a circle Of decl.

, and P’

XM, a circle

«of latitude

,

and join Aand X by an arc of a great circle.

In the figure are given, AR the right ascension, XR thedecl. , and MAR the obliquity, to findAM the longitude, andXM the latitude.

2 16 SOLUTIONS OF PROBLEMS.

included angle, tofind the third sideAR=7° 29’16

the distance between the two places in nautical miles.Or thus since AP=PB , a perpendicular drawn from P

upon AB will bisect the angle P and are AB. Suppose ‘

in

the figure an arc AD to be drawn perpendicular to AB thenin the right-angled . triangle PDA are given AP= 45 and

angle APD=5°

to find AD=half the distance AB.

Rule XIII sin. AD=Sin . AP sin . APD.

PROB. 135 (Fig.

Let UV be the arc of the equator intercepted between

the meridians PB and PV passing through B and A, the

two places ; join AB. Then in the triangle BPA are given,PA=90° — 50

°

and the included angle P=57° difference of longitude

between the places.

Whence (by Rule IX. or X. ) the third side AB may befound=99° 9 ’

or 5 949 8 nauticalmiles.

60nautical miles=69 °05 geographicalmiles

hence to express the above distance in geographical miles wehave 60 w=6847° 2 geographical or English miles.

Great Circle Sailing.

(l . ) Find the shortest distance between Liverpool andNew York.

lat. Liverpool 5 3° 25’ N. long. Liverpool 2° 59

’W.

New York 40 42 N. New York 73 5 9W.

Let PL,PY be the

meridians of Liverpooland New York, and LYan arc of a great circle

passing through the two

places then in the ephorical triangle PYL are givenPL= 9O°— 53

° PY=9O°— 40° and the includedangle P=71

°

, the difference Of longitude between the two

IN NAUTICAL ASTRONOMY. 2 17

places, to find (by Rifle IX. orX. ) the arc. YL=47° 5 2’

2872 nautical miles.

Take a

point A on the

arc YL,distant

fromLiverpool,L

,

872 miles ; takeanother point B ,

distant from A

1000miles ; thenfrom B to NewYork, the distance BY= 1000 miles. TO find the latitudeand longitude of each of the pointsA and B .

In the triangle PYL are given the three sides,tofind

(by Rule VIII .) the angle L=75°8

’

To find PA,and thence the latitude ofA.

In the triangle PLAare given . the Sides AL or

14° PL

, and the angle L, just found, to find PA (thecolatitude of A)=35

°and thence the lat. of A

5 4° 39’ N.

To find angle APL,and thence the longitude of A.

In the same triangle APL are given the three sides, tofind the angle APL=°4

°47

’

angle APL= 24° 47 15

”W'

.

long. of Liverpool= 2 5 9 0W.

long. Of point A= 27 46 1 5 W.

To find PB, and thence the latitude Of B .

In the triangle PBL are given the sides BL or

31° PL

, and the angle L, to find PB (the colat. OfB)and thence the lat. Of B= 50° N.

TO find the angle BPL, and thence the longitudeof B.

In the same triangle BPL are given the three Sides, tofind the angle BPL= 5 l

°


angle BPL : 5 1°15'w.

long. ofLiverpool: 2 5 9 W .

long. Of point 9 W.

The points A and B being thus found on the line of

shortest distance, the course and distance from Liverpool toA

,thence from A to B ,

and thence from B toNew York,may be found by the common rule ofNavigation .

The course and distance from Liverpool toNew York byMercator’

s sailing are S. 75 ° 10’30

” W.,2982 miles, and

the sum of the distances described by the ship sailing on the

three loops, as above, is 2884, so that the distance saved byaltering the course twice is about 98 miles.

'

The above method Of finding the length of the arc of a

great circle passing through two places on the surface of the

earth, and of determining a certain number of convenientpoints thereon, to which the ship may be directed, so as tokeep her as near to a great circle as possible, is direct and

general. By dropping a perpendicular from P, or by meansOf special tables, the computation may be shortened, but onlywith the disadvantage Of introducing a distinction of cases

,

and rendering the problem complicated. As the attempt at

great circle sailing can only be of rare occurrence, it does notseem advisable to give formal rules for the purpose, nor toburden our books

'

with more tables than are already in use.

By taking a greater number of points than A and'

B on

the arc YL, so as to bring the ship oftener to the line of

shortest distance, the sum Of the distances sailed will approximate nearer to the shortest distance. It will be seen,

however, that even on the above assumption of only alteringthe course once in about 1000miles, the sum Of the dis

tances run exceeds by a very few miles the shortest distance,and that the absolute saving in distance between the two

points Liverpool and New York is on this supposition nearly100miles.

PROB. 136 (Fig.

Let AQ represent the celestial equator, A the first noint


or 1°

Through Z and S draw the meridians ZQ, SQ,

meeting in pole P ; then the angle SZQ is the initial course

and the angle PSZ,or its opposite angle, is the course

of the ship at S . To find thisangle, we have in the trianglePZS the two Sides (PZ colatitude, and ZS the distance run),and the included angle PZS= 135

°. Hence the angle at

S may be found by Rule XI.

PROB. 1 40 (Figs. 86,

In this problem it is not stated whether B is to the east

or west ofA. As the longitude of C depends on its position with respect to B as well as to A,

there will be twosolutions. (See figs. 86,

LetA, B,C,be the three places ; then we have given ,

PA (the colat . of PB (colat. of andAB :

AC to find CP,the colat. of C, and angle CPA,

the difference of longitude between A and C,whence the

longitude Of C may be found.

In triangle ABP are given the three Sides,to find

the angle BAP= 5 1°

In triangle BAC are given the three sides, to find

the angle BAC= 61° l’

30 hence the angle PAC

9° 13’

(3) In the triangle PAC are given , the two sides PA

and AC, and included angle PAC,to find angle APC (the

difference Of longitude between A and C)= 6° 13

'

and

the Side PC (the 23’

If C is to the east ofA (as in fig.

then the long. of C= long. A+ 6°13

’

30”

21° 13’

30 E.

If C is to the west ofA (as in fig.

then the long. of C= long. A— 6°13

’46

’

0 E.

PROB . 141 (Fig.

Let T and R represent the places of the stars, C that of

the comet,AEQ the ecliptic, and P the pole of ecliptic.

Through the heavenly bodies T, C, R, draw the circles of

IN NAUTICAL ASTRONOMY. 22 1

latitude ; then CE=lat. OI comet, and AE the longitude

(A being the first point of Aries) .

(l .) In triangle TPIR are given the sidelat.= 95° 28

’

the side P1R=90° 32

’

and the included angle TP1R=difference of longitudes

80°3

’

to find the third side TR=80° 8’

In same triangle TP lR are given the three Sides, tofind the angle PITR=88

°34

’

In triangle RTC are given the three sides, to find

angle RTC=39°32

’

Hence the angle P lTC is known, being the sum Of the

last two results= l 28° 7

In triangle PlTC are given TPI=95° 28’

TC

= 40°

and included angle P,TC : 128°

to find PIC :

1 18° 0’

1 5”

Hence CE the lat.= 2 8° 0

’15 S.

Lastly, in the same triangle PITC are given the threesides, to find the angle TPIG (the difference of long. be

tween comet and Aldebaran)=35°6

’ Hence the longitude=67° 12 ’

l 5”

+ 35° 6

’40 102

°18

’

PROB. 142.

Let the fig. be

projected on the planeofthemerid. PEP

IQ,

.Z the zenith of spectator

,and Elk the

plane Of the horizon .

Take HP=54°

then P is the placeOf the elevated pole

(Trig. Part I. p. 126)through P draw the

diameter PP,,

and

EQ at right angles toit . Then PP1 is theaxis Of the heavens

,celestial

222 SOLUTIONS or PROB LEMS

equator. Draw dd' parallel to the equator, at the distancefrom it equal to the sun ’

s declination 8° 30’ N.

,and let S be

the sun’

s place when twilight commences ; draw ZS, PS,

arcs of great circles ; then S is 18° below the horizon , and

ZS= 108° . In the triangle ZPS are given ,ZS= 108

°

,co

latitude ZP=35° and polar distance PS=81° to

find the hour- angle ZPS= 9h. 14m. 1 63. Hence twilight

begins at 2h. 45m. 44s. A.M. , and ends at 9h. 14m. 168. R M.

(the declination being supposed to remain unaltered) .

PROB. 143 (Fig.

(lst solution .) Let X and Y be the places of the sun at

the times of the observations ; PX,PY, the polar distances,

each= 66° and ZX, ZY,the zenith distances= 50° 10"

and 61°20

’

and angleXPY= 1h . 30m. (the intervalbetweenthe observations) . It is required to find PZ the colatitude,and thence the latitude. Draw XY an arc of a great circle.

(l .) In triangle PXY, given PX,PY

,each 66° and

included angle XPY= lh. 30m.,to find XY= 20° 37

’

In triangle PXY, given the three sidesh to findangle

PYX= 85 ° 28'

In triangle ZYX, given the three sides,to find angle

ZYK=5 1°

Hence PYZ (= PYX 47'30

Lastly. In triangle PYZ are given, PY , ZY, and in

cluded angle PYZ ,to find the third side PZ= 30° 42

’

the

colat. latitude= 90°— 30°

18’ N.

(2d solution [fig. Let X and Y be the places of thesun at the times ofobservation PX,PY,

the polardistances

(which in this solution is always supposed to be equal) .Bisect XY in M , and join PM,

ZM ; then PM is at rightangles to XY , and therefore the angle PMZ is the complement ofZMX draw ZE perpendicular to PM .

Let alav2represent the altitudes at X and Y

,72 the half

interval MPY between observations, d the decl. , and l the .

latitude.


Lastly, in the right- angled triangle PZE,

cos. PZ or sin . lz cos. u cos. v.

From this solution is deduced the rule in Navigation ,

Part I. p . 1 56,for finding the latitude by double altitude ;

Observing that in finding y we must take the supplement ofthe are given by the tables when the lat. and declination are

of different names.

From solution 1 is derived the rule for finding the latitude by double altitude given in Navigation , Part I. p . 142 .

This method is perfectly general, since it may be applied tothe same or different heavenly bodies

,taken at the same or

different times. The rule Obtained from the second solutionis considered more concise when the same heavenly body isobserved, if its declination does not alter in the interval between the observations ; as when two altitudes of a star aretaken. It will also give a near approximate latitude fromtwo altitudes of the sun the declination in this case beingsupposed to remain invariable

,and to be taken out Of the

Nautical Almanac for the middle time between the observations.

PROB . 144 (Fig.

Proceeding in a similar manner as in solution 1 of lastproblem

,we find KY= O2

° PYX= 1 5 5 ° ZYX=

89°57

’

PYZ=65 ° PZ=63° 36’

whence lati

tude=26° 23’

7 N.

PROB . 145 a (Fig.

Let A and B be the two places in the same latitude, PA,

PB,their meridians. Draw PD at right angles toAB,

the

great circle passing through the two places then PD bisectsthe angle APB , since AP=BP and D is the highest pointreached by the ship sailing on a great circle fromA to B .

Produce also themeridians to the equator UV,and draw the

parallel of latitude ADIB .

III right-angled triangle APD, given PA= 5G°

and

angleAFD=half difference of longitude between A and B


68°

to findAD= 5O° 23’

45 and PD= 29° 6'

l'

at. of

D=GO° andAB= 2 AD= 100° 47 or 60475 nau

tical miles, the distance between A and B on a great circle.

To find the distance AD,B on a parallel of latitude,

we have (Trig. Part II. p.

AD1B=UV cos. AU= difi

'

. long. x cos. lat.

8170cos. 33°5 l

’=6785 miles.

whence difference on the two circles= 737°5 nautical miles.

PROB. 146 (Fig.

LetA and B represent the two places, PA and PB theirmeridians. Draw the perpendicular PD,

then the latitude ofD is the highest attained by ship .

(l .) In triangle APB, given PA= 5G°

PB=34° and

angle APE= 140°

to find the angle PAB=20°5 9

’

In right-angled trianglePAD, given PA= 56° 9

’and

angle PAB= °O° 59

’45 to find PD= 17° 18’

45 lat. of

D= 72° 41’

15

PROB. 147 (Fig.

Project this figure on the plane of the celestial meridianO f the place. Let P be the pole

,EQ the celestial equator,

Z the zenith,S the place of the sun below the horizon Hh,

PS a circle Of declination , and ZS a circle of altitude.

(l .) In triangle ZPS there are given,decl.= 100° PZ the colat.=39° 12 ’

and hour-angle P7h. tofind the third side SZ= 107° 24 whence 107° 24'

— 9O = 17°24

’the sun

’

s depression .

In the same triangle the three Sides are now known,henceangle PZS, the azimuth,may be found=N. 84

° 53'

W.

PROB . 148 (Fig.

Let X and Y represent the two stars on the same circleO f altitude. It is required to find the azimuth Of X and Y

,

or the angle PZX .

In trianglePXY,given PY=GI°35'34


and included angle XPY=3h. 8m. 27s. (the difference of

right ascensions), to find angle PXZ= 65° 46

'1 5

”

In triangle PXZ , given PX, PZ (the and

angle PXZ , to find PZX= 104° whence its supplement— S. 75° 5

'W. the bearing required.

PROB . 149 (Fig.

Let M and X represent the two bodies on the same vertical circle ZM ; it is required to find PZ

,the colat. of the

place, and thence the latitude.

In triangle PXM, given PX=77° 14’ PM :

9 1°

42'

and included angle XPM= 2h. 36m. 12s. (thedifference ofright ascensions), to findanglePMZ= 68

°23

’

In triangle PMZ , given PM and ZM the moon’s zen .

dist. and included anglePMZ ,tofind colat. PZ=70° 4

’

and therefore lat.= 19° 5 5 ’

38” N.

PROB. 1 50a (Fig. 7

Suppose the heavenly body to rise at S and to pass the.

prime vertical at X,then PS=PX the star’

s polar distanceand EX=star

’

s altitude when due east= 20°. Let a= alt. atX ; m=amplitude ES= 1 1

° w= lat. of spectator, and ydecl. of star.

Then PS=PX= 90° — x,and ZX=90

°

In quadrantal triangle PZS (Rule XIV.)sin . y= oos. a: sin . m

In right-angled triangle PZX (Rule XIII. )sin . y=sin . x sin. a

equating (1) and sin. to sin . a cos. a: sin . m

SIII. m

sm . a

sin . 1 1°

cosec. 20°

a=29° 42 ’ N.

tan . x=sin. m cosec. a

PROB. 15 1 a.

Describe fig. to Prob. 107 take a distance EX=10° on

“


PROB . 1 53 a (Fig p.

Let V and D be the places of the body at 6 o’

clock and

when setting ; join PD then in the triangles PDZ and

PVZ are given, DIV the alt. at 6 o

’clock

,and DZP the

complement of the amplitude, to find the latitude and decl.

Let a=alt. DIV at 6 O

’

clock,m= amplitude WD,

a:

lat.=90°— PZ, y=decl.= complement of PV or PD .

Then in triangle PDZ sin . g= cos. a: sin. m

PVZ sin . a=sin. y sin . a:

Multiplying, and cancelling,

sin. a=sin . a: cos. a: sin . m i sin a: sin . m

sin . 2a: 2 sin. a cosec. m

whence 2x= 84°, or 96

°

(the suppl. ) and x=42°, or

Substituting these values Of x in we have y=22°20

’

or 20°

PROB. 1 54 a (Fig. p.

Let V and D be the places of the body at 6 o’clock andwhen setting join PD .

Let h=hour-angle DPZ when setting, a=alt. DIV at

'6 o’clock, — PZ

, ofPD or PV

Then (Rules XIV and XIII).

sin . a=sin . y sin a:

sin . sin. a: sin . aFrom (1 ) cos. b

y

cos. y cos. a: cos. y cos. a:

cos. y cos. x cos. h — sin . a

and cos. y . cos. a: — sin . a sec. 71.

But by (2) sin . y sin . a=sin . a

Adding and subtracting, we have

cos. as. cos. y+ sin . ac . sin. y : -sin . a . ( l — sec. h)cos. as. cos. y

— sin . a . sin . y=sin . a . (1 + sec. h)or cos. (ax a (1 — sec. h)

cos. - sin . a (1+ sec. h)


To reduce (3) and (4) to a logarithmic form.

cos. (x —

coi. h)sin.

— 1)

- 2 sin . a sin g; sec. 71.

- 2 sin . a coal;3 (6)

From equations (5 ) and (6) the values of r —y and art- y

“

may be determined, and thence a: and y.

PROB. 15 5 a (Fig. p.

Let k=hour-angle at setting, as at D,hl=hour-angle at

Y on prime vertical, x=lat. , y=decl.

Then cos. 7: - tan . a: tan . y

cos. h tan. a:

cos. h,

cot. a:Dividing (1) by

or tanfia: — cos. h sec. 71 which equation determines thevalue of £6 ; and multiplying (1 ) and (2) together, we havetan .

2y

— cos. h cos. kl ,whence the declination can be

PROB. 15 6 a (Fig. p .

Let h=hour-angle at setting, h,=hour-angle when west,t==interval i=h— h

,,d=decl a=1atitude.

cos. t= eos. (h— hl) cos. h cos. h

,+ sin . h sin . b,

By Rule XIV. cos. h= — tan . a: tan . d

XIII. cos. h,= cot. a: tan . d

Multiplying (2) and . cos. h cos. III

or 2 cos. h cos. h1

Subtracting equation (1 ) from this equation,cos. h cos. hl

— sin . h . sin hl

— cos. t— 2 tan .

2 d

— cos. cos. t+ 2 tan.2 d

2 30 SOLUTIONS or PROBLEMS

This equation determines the value of [t+kv the sum Of

the hour-angles ; having this, and the difference, the angles

h and 721 ,and thence the latitude at

,may be found. The

operation is as followsTo find value of — cos. (h+h1) cos. t+ 2 tan .

2 d.

tab. log. cos. t= 9 °864304 log. cos. t= l °864304

cos. t= '

73165 .

To find value of 2 tan .

2 d.

log. (2 tan.

2 d)= log. log. tan. d

2 tan .

2 d= °26495

cos.

:angle corresponding to log. of this in tables= 18"ll

Now,since the cos. (h -Hz

l) is negative, the value of71,+ hlmust be either 18

m5 4

'

or

that is h+hl= l l

h41

“6'

or

and since h —h1

2 5 1 54 2 5 1 54

2h= 14 33 O or 15 10 48

and h : 7 16 30 7 35 24

With the first value of h the lat.= 42°

other. 48 nearly.

PROB. l5 7a (Fig. p .

Let Z=angle PZV, the azimuth at 6 o’

clock ; m=amplit udeWD,

d=decl.,l=lat. then

Rule XIV sin. d=cos. l sin. m

XIII, cos. l= cot. d . cot. Z

Multiplying, sin . cl cos. I cos. I . Sin . m cot. d cot. Z

sin .

2 d=sin . m cot. Z cos. d

Or 1 cos.

2 d=ein . m cot. Z cos. d

whence cos.

2 d sin . m . cot . Z cos. d=1

C ompleting square, and extracting the root,cos. d: sin . m . cot. Z+ § J sm.

2m cot.9 z+ 4

tab. log. cos. d= 9 °972984 d=20°

and thence Imay be found.


whence 2l+ a= 29° or 1 50° 47’

(its supplement). Tak~

ing this last value, 2l= 94° 47

'Z=47° 23

’

30”

To avoid the ambiguity arising from obtaining the valueof 2l+ a in terms of the sine

,let a= 90° — a the zenith dist

then d= l— z

cos. h - tan . d tan . 1 — tan. (l— z) tan . 1

— 2 sin . (l — z) sin . 2 cos. (2l — z) — cos. 2

2 cos. (l— z) cos. l_cos. (2l— z)+ cos z

cos. h+ l cos. (2l— z)or cos. (2Z— z) cos. z cot 2

cos. h— l - cos. 2

from which equation the value of Imaybe obtained free fromambiguity.

PROB. 160a (Fig. p.

Let k=hour-angle at setting, h,=hour-angle when west.

By Rule XIV cos. h — tan. 2 ta‘

n . dXIII cos. b

l=cot. l tan . d

adding, cos. h+ cos. l— tan . l) tan . d

subtracting, cos. h cos. I),

— (tan . l+ cot. l) tan . d.

cos. h+ cos. h,

cot. l— tan . ldwldmg’

cos. h — cos. h,

tan . l+ 00f~5

cos. h+ cos. It,

cot. Z— tan . 1

— (cos. h — cos. 717) tan . l+ cot. l

2 cos. cos §l (h h

,) 1 — tan .

2l

2 sm . gown) . gout — h) tan .

2l+ 1

or cot. cot. 501. — h,) cos. 2 l

cot. cos. 2 l . tan . fl it — h)

which equation determinesMk andwith&(h h.) already

known, the angles h and h,may be found, and thence by

Rule XIII. we can find d.

—cos. 2 l

PROB. 161 a (Fig.

Suppose the heavenly body to rise at S, and to describethe parallel of declination SX. Let a circle of altitude ZOtouch the parallel SX at X then X is the point where the


azimuth is the greatest ; the azimuth increasing from the

point S,where the heavenly body rose, to X, and beginning

to decrease after passing X. The angle PXZ is evidently aright angle ; whence the required azimuth, hour, and altitudemay all be found from the right-angled triangle PZX.

(L) TO find greatest azimuth PZX.

cos. d=eos. 2 . Sin . az . sin . az cos. d . sec. 1

To find the hour-angle ZPX, Or Iz.

cos. [a= cot. d tan . l.

To find altitude X0 or a.

sin. Z=sin . d sin . a sin. a=sin . l cosec. d

PROB . 162 a (Fig.

The preceding problem furnishes an explanation of the

fact, that when the sun’

s declination is greater than the latitude of a place, and Of the same name with it, the shadow Of

an upright object on a horizontal plane goes backward eachday during a certain period, which may be computed, as wellas the are through which it goes back.

At the point X the sun appears stationary in azimuth .

PZS is its bearing when rising, PZO its bearing at point X.

The angle SZO or are SO will therefore denote the numberO f degrees the bearing has increased from sun rising, or thatthe shadow of the Object has gone back ; and the periodduring which the retrograde motion has taken place will bemeasured by the angle SPX.

To find bearing PZS at rising.

In triangle PZS, sin . a=cos. l cos. PZS

whence PZS= 64° 5 5’

30

By last prob. PZX= 77 28 0

shadow goes back= l2 32 30

234 SOLUT IONS OF PROBLEMS

To find hourangle ZPS at rising.

In triangle PZS, cos. ZPS — tan. 1 tan . d

ZPS= 6h. 36m. 228.

by last prob. ZPX=2 12 7

time ofshadow’

s going back=4 24 15

PROB. 163 a (Fig.

(lst'

solution. ) Let S and M be the respective places Ofthe sun and moon at the time Of observation ; d and — d the

decl. of sun and moon respectively ; that of the moon beingtaken negatively, since the bodies are on different sides of theequator ; h and h, theirhour-angles.

Then h 5 3m. 428.

In triangle ZPS, cos. I: — tan . Z tan . d

In triangle ZPM,cos. b

,

— tan . 1 tan .-d

,)

cos. b,

tan . d,

cos. h — tan . d

cos. h,+ cos. h tan . d

,

- tan . (1

cos. h,

- cOS. h—

tan . d,+ tan. d

2 cos.Mk,+h) cos. g ig— h) sin. d, cos.a— cos. d sin . (I

— h) sin . d,coe.d — cos. d

,sin . at

an (d, - d)sin.

cot. — sin . (d,— d) cosec. tan. %(hI— h)sin. (d, d) cosec. (d, d) tan .&(h h

,)h)=5h. 5 3m. 29s.

§(h 5 6 5 1

h= 6 50 20

hence app. time=l 2h.— 6h. 50m. 208.=5h. 9m. 408. A.M.

To find lat. Z. cos. h — tan. 1 . tan. d

or tan . l=cos. h cot. d l= 50° 18’20

” N.

dividing (2) by

or — cot. fl hl+h) cot. 504— 71)


To find the angle PAO.

Draw OE perpendiculartoAD thenAE=EO=§AD150feet.

OP so 6and tan - PAC)

Ar3=15w 2 at“?

whence angle PAO= 15° 47

’the sun

’

s altitude.

PROB . 166 a (Fig.

Let ZQH represent the celestialmeridian, H the horizon,Z the zenith, Q the equator, S

’

and S the places of the sunon the longest and shortest days.

Then ZQ= lat. and QH= 90°

— x

28'

Let NG (the shadow of Object PNwhen sun isat

then NC (its shadow when at

Now angle NPC or are ZS =az+dand angle NPG or are ZS

’= at — d

In triangle NPC 7y=PN tan. NPC= PN tan.

NPG y=PN tan. NPG=PN tan . (a' - d)

tan . (a+d) sin . 2a:+ sin. 2ddIv1d1ng 7_

tan . (az— d) sin . 2a:— sin . 227

8 4 SIn . 2a: 4

6 3 sin . 2dor sm. 2a'

3sm. 2d

whence ac= 38° 27 40 =lat.

PROB. 167 a (Fig.

Let NG=a3, NC=y, l=1at. , d : decl.then w==PN tan . (l— d), y=PN tan. (l+d)

y tan . 2l+ sin . 2d

sin . 1+ -

2L 3

Sin.90° — sin . 30° li f i

or y z a'

IN NAUTICAL ASTRONOMY.

PROB. 168 a.

In fig. 98, if S’

be the place of the sun, and NG the

shadow of object PN,then

P

Ng— gz tan. PGN=tan. sun’s alt

log. tan . alt.=9‘77815 1 alt.=30°

Then in triangle PZX (fig. 7 the three sides are given,viz. PZ= 56° PX=79° andZX=5 9° to computethe hour-angle ZPX, or k ; hence h=3h. 58m. 4s.

PROB . 169 a (Fig.

LetAB represent the shadow of perpendicular stick AD,

and AC its shadow when the stick is placed in the positionAT perpendicular to ray SC, in which position it will manifestly throw the greatest shadow. Now the distance of S isso great compared with BC that the angle BSC has no as

signable magnitude, and the rays SB,SC may therefore be

considered parallel, or the angle ABD=angle ACT=sun’s

altitude required.

In triangle ABD AD : AB tan. alt.=4 tan . alt.

In ATC AT=AD=AC sin . alt.= 5 sin . alt.

5 sin. alt.=4 tan . alt.=4 sin . alt. sec. alt.

sec.

log. sec. alt.= 36° 52’15

PROB. 170a (Fig.

LetA be the Observer, B the Shadow of cloud, C and Dthe place of cloud and sun . Since the distance AB willsubtend no sensible angle at the sun, the raysAD and BD

may be considered parallel, and angle CBE=DAB=sun’s

and the angle of elevation of cloud= 20°. Therefore ACB=2°.

BC SIn . 20°

In tri. ABCH i sin . 2

0


In right-angled triangle CBE CE=CE sin. 22°

CE=AB sin . 20°

sin. 22°

cosec.

PROB . 171 a (Fig.

Let h denote the hour-angle of the sun at rising, or halfthe length of the day and since the meridian altitude is lessthan the latitude, the declination must be of a different nameto latitude, or south.

In quadrantal trianglePZS, cos. h — tan. 1 cot.(90° d)

-tan . l tan . d= tan . 01,since tan. Z=tan. 45

°= l .

cos. b=tan . (z — l) if z=zen . dist.tan . z — tan . 1 tan . z— l

l + tan . z tan . l tan . s+ l

l + cos

tan . a=cot. alt.=cot.

l4hence cot J 3 and tan

'3/3

PROB. l 72 a.

Let z=zenith dist. s=2l

and (fig. 66) z= l+d 2l= l+d or l=d.

In triangle PZS (fig. cos. h — tan . 1 cot. (90°

tan . l . tan . d= tan .

2l

tan .

2 l=cos. 7h., whence l= 26

°

PROB . 173 a (Fig. p.

Suppose D2the place where the star rose draw PD

2and

ZD,Zlet h=hour-angle, and z=meridian zenith distance

,

l=lat. , and d=decl. Thencos. h= — tan . l tan . d — tan. d since l= 45 °

but d=l— z cos. h — tan. (l— z)tan . 1 tan . 2 tan .

'

z — l


Now (Trig. PartII. p. 80)UV=AB sec.AU=16’17-3 . sec. at

9

12-

2860. d= lm. 10°22s.

(this is independent of the sun’s apparent motion from west

to east).

or are UV (in time) :

PROB. 176 a (Fig.

Let A be the first station Of ship,B the second, and C thepoint ofland.

(1) In triangleABC, given AE=Gmiles, angle CAB (thedifference of bearing between C and B)= 2 points, and angleACB=4§ points, to find AC.

(5) Draw AD due east, and CD at right angles to it, thenCD= difi

'

erence of latitude betweenAand C,and (considering

ADC as a right-angled plane triangle) CD=AC sin . CAD— AO sin. 4 points= 5 ’

15”

lat. of C=5 9° 15 = 5 9° l l’15

also (the triangle ADC being isosceles) AD= 5’

1 5

To find the difference of longitude between A and C, wehave (Nav. p. 43) AD=difi

'

. long. cos. lat. A, whence difi'

.

long.= 10

’13 E.

long. of C=6° l5’

+ 10’

l3"=6° 25

’13 E.

PROB. 177 a (Fig.

Let S and M be the places of sun rising on the longestand shortest days,

It =hour-angle of sun at S,

MI

then 2 (h in length of day.

In triangle ZPS cos. h — tan . d tan . l

. .ZPM cos. kl: + tan. d tan . Z

cos. b — cos. Iz,=cos. (12h.

— h) h= 12 — h,

or 71. h,12 2h

,difi

'

erence required.

Now in cos. h,=tan . d . tan . I=tan . d

,since l=45°.

IN NAUTICAL ASTRONOMY . 241

log. tan . d 61 1 =log. cos. h,

h,= 4h. 17m. 5s. ,

and 2h,=8h. 34m. 10s.

h — 2h,) =3h. 25m. 50s.

and difference in length of days=6h. 5 1m. 40s.

PROB. 178 a (Fig. 97)Let h=hour-angle of the sun at S

,

h

triangle ZPS cos.— tan . d . tan. 7 (1)

ZPM cos. h,= tan . d . tan . l (2)

cos.— cos. h

,=cos. (1 2 — h

,)h,+ 3= l2

— h,or h

,= 4h. 30m. ; whenceby l=41° 24

'

PROB. 179 a.

Let h and 71,be the hour-angles then h

d and d,the decl. at these times, viz. 20

°and 10°

then cos. b. — tan . 1 . tan . d .

cos. h,

— tan . 1 . tan. d,

cos. h tan . d cos. h+ cos. tan . d+ tan . d

cos. h,

tan . d,

cos. h— cos. h,tan . d— tan . d

,

2 cos. cos. Mir— h) Sin .

2 sin . + h,) sin . (d— d,)

sin . 30°

— cot. cot. §(h 100

— cot. 7°

sin. 30°

cosec. 10°

whence 23m. 3s.

and &(h— hl)=0 30 0

5 3 3

with this value of h, the lat. (by 27 N.

PROB. 180a (Fig.

Let h be the hour-angle of sun at S,

M h= 3k,


then cos. 1: — tan . i tan. d

cos. h,= + tan. l tan. d

cos. 71 — cos. h,

—cos. (12h.

— h,)

h= 12 — h,

or 3h,= l2 — h

,4h

4= 12

and h,=3 hours whence, by l= 58° 27 N.

PROB. 181 a.

3Let h and h

,be the hour-angles, then h, 5

k

and cos. h — tan . d tan . l, cos. 7c,

= tan . d tan . 1

cos. Ii — cos. hl=cos. (12 — k

,) h= l2 — h,=12 — §Iz

§5

IE 12 and li -6

8

07h. 3om.

with this value of h the lat.=41° 24 ’ N.

PROB. 182 (Fig.

Let A be the place sailed from,B the place arrived at ;

the arcAA,=arc and BC=b EQ the arc of equa

tor between meridians OfA andA,.

Let x= lat. ofA; then x— a=lat. ofB. Trig. Part II. p. 80,

CB 5 AA,

a

lflQ—

EQ— cos' BQ cos. (ar— a) , EQ

"

Ecos. AE

b 5 cos. (cc— a)008. x

cos. (as— a) a cos. 93

cos. a; cos. a+ sin . a: sin. acos. a+ tan. a: 8111. a

cos. a:

cos. a

tan . a:an. a

b= 150, a=lo0. tan . a:

log. cos. 1°40

'

19 99816

tan. xsin. 1

°40

’

Calculation.

cos. 1° 40

’

sin. 1°40

'

cos. 1° 40

’

5006 cosec. 1°40

’

2 44 SOLUTIONS OF PROBLEMS IN NAUTICAL ASTRONOMY.

‘

To find u s.

7° 5 1’0"

0004089

tr. alt 10 36 24 0007487

2 45 24 49 2641 1

pol. dist 1 12 24 0 4 912419

1 15 9 24 98 50397

109 38 36 PZS 1 14°

and OZX 95 39 0

bearing of point N. 19 0 30W.

In the preceding example suppose O to be the summit of a mountain whose elevation is it is required tofind the true bearing of the point 0. In the preceding cal

culation,to find angle OZX we must substitute the zenith

distance of the point 0 for the quadrantal are OZ inlast example . We have then three Sidesgiven of the triangleO ZX, to find as before the angle OZX.

Calculation.

ZO 80°0

’

0”

79 18 46

0 41 14

CX . . 95 32 5 5

96 14 9

94 5 1 41

bearing

THE END.

Losnov

consort AN» sous, PRINTERS , PANCRAS ROAD,N.W.

0006649

0007594

4871860

4 867153

97 5325 6

OZX 97° 39’

andPZS 1 14 39

ofmountain N. 17 0W.

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