Spherical Trigonometry £171? 1 E Me Hi Qeniiegez 31th $5

I N t h e th ird e dit ion I h ave made some addition s wh ich I

h ope will be foun d valuable . I h ave con side rably e n larg e d

t h e discussion on t h e con n e xion of Formulae in Plan e an d

Sph e rical Trig on ome try ; so as t o in clude an accoun t of t h e

prope rtie s in Sph e rical Trig on ome try wh ich are an alog ous t o

t h ose of th e Nin e Poin ts Circle in Plan e Ge om e try. Th e

mode of in ve stig at ion is more e leme n tary th an th ose h ith e rtoemploye d ; an d pe rh aps som e of t h e r e sults are n ew. Th e

fourte e n th Ch apt e r is almost e n tir e ly orig in al, an d may de

se rve atte n t ion from t h e n ature of t h e proposition s t h em

se lve s an d of th e demon stration s wh ich are g ive n .

CAMBRIDGE ,

July, 1 87 1 .

CONT E NTS .

Gre at an d Small Circle sSph e rical Tr ian gle sSph e rical Ge om e try

Re lation s be twe e n th e Trigon ome trical Fun ction s of. th e Side san d t h e An gle s of a Sph e rical Trian gle

Solut ion of Righ t -an gle d Trian gle sSolution of Oblique -an gle d Trian gle sCircumscribe d an d In scribe d Circle sAr e a of a Sph e rical Trian gle . Sph e rical Exce ssOn ce rtain approximat e FormulaeG e ode tical ope ration sOn small variat ion s in th e part s of a Sphe ri cal Trian gleOn th e con n e xion of Formula in Plan e an d Sph e rical Trigo

n ome try

Polyh e dron sArcs drawn to fixe d poin t s on t h e Sur face of a Sph e re

Misce llan e ous Proposition sNume rical Solut ion of Sph e rical Trian gle s

SPHE R I CAL T GONOME TRY .

I. GRE AT AND SMALL CIRCLE S.

l . A SPHE RE is a solid boun ded by a surface every poin t ofwh i ch is equally distan t from a fixed poin t wh i ch is called the

c e n tr e of the sphere. Th e straig ht lin e which join s an y poin t of

the surface with the cen tre is called a radius. A straig ht lin e

drawn throug h the cen tre an d termin ated both ways by the surface

is called a diame te r .

2 . Th e se ction of th e surface of a sp h e r e made by

is a circle .

Le t AB be the section of t h e surface of a sph ere made by an y

plan e, 0 the cen tre of the sphere. Draw 00 perpen di cular to t h e

plan e ; take an y poin t D in the section an d join 0D, CD . Sin ce

00 is perpen dicular to the plan e, the an g le 001 ) is a rig ht an g le 5therefore 01) J(0D

2Now 0 an d 0 ar e fixed poin t s, so

that 00 is con stan t ; an d OD is con stan t,bein g t h e radius of t h e

T . S . T.

2 GREAT AND SMALL CIRCLES.

”

sphere hen ce 0D is con stan t, Thus all poin ts in the plan e sectionar e equally distan t from the fixed poin t 0 therefore the sectionis a circle of which 0 is t h e cen tre.

3 . Th e section of the surface of a sph ere by a plan e is called

a g r e at cir cle if the plan e passes throug h the cen tre of the Sphere,an d a small cir cle if the plan e does n ot pass throug h the cen tre of

the sphere. Thus the radius of a g reat circle is equal t o theradius of the sphere .

4 . Th roug h the cen tre of a sphere an d an y two poin ts on thesurface a plan e can be drawn an d on ly on e plan e can be drawn ,

except when the two poin t s ar e the extremities of a diameter of

the sphere,an d then an in fin ite n umber of such plan es can be

drawn . Hen ce on ly on e g reat cir cle can be drawn thr oug h two

g iven poin t s on the surface of a sphere, except when the poin ts ar et h e ext remi t ies of a diameter of t h e sphere . Wh en on ly on e g reat

circle can be drawn throug h two g iven poin ts , the g reat circle isun equally divided at t h e two poin ts we shall for brevity speak of

t h e short er of the two arcs as th e arc of a g reat circle join in g the

two poin ts .

5 . Th e axis of an y circle of a sphere is that diam e t e r ~of th e

sphere which is perpen dicular t o the plan e of the circle ; the e x

t r emi t ie s of the axis ar e called the p ole s of the circle. Th e poles

of a g reat circle ar e equally distan t from t h e plan e of t h e circle.

Th e poles of a small circle ar e n ot equally distan t from the plan eof t h e circle they may be called respectively t h e n e are r an d fur

th e r pole 5 sometimes t h e n ear er pole is for brevity called th e pole.

6 . A p ole of a cir cle is equally dis tan t from e ve ry p oin t of th e

circumfe r e n ce of th e circle .

Le t 0 be the cen tre of the sphere, AB an y circle of the sph ere,0 the cen tre of t h e circle, P an d P

' the poles of t h e cir cle. Take

an y poin t D in the cir cumferen ce of t h e circle join 00 ,0D,

PD .

Then PD J(P02+CD

2

) an d P0 an d CD ar e con st an t , therefore

PD is con stan t . Suppose a g reat circle to pass throug h the poin ts

P an d D 5 then the chord PD is con stan t, an d therefore the arc of

GRE AT AND SMALL CI RCLE S. 3

a g reat circle in tercepted between P an d D is con stan t for all

position s of D on t h e circle AB.

Thus the distan ce of a pole of a circle from every poin t of t h e

circumferen ce of the cir cle is con stan t,wh ether that distan ce be

measured by the straig ht lin e join in g t h e poin ts, or by the arc of

a g r eat circle in tercepted between th e poin t s.

7 . Th e ar e of a g re a t circle wh ich is drawn from a p ole of a

g r e at circle to an y p oin t in its cir cumf e r e n ce is aquadran t .

Le t P be a pole of the great circle AJ30 th en t h e arc PA is a

For let 0 be t h e cen tre of t h e sphere,an d draw P0. Then

P0 is at r ig ht an g les to the plan e AI 90, because P is the pole of

ABO, th erefore POA is a rig h t an gle, an d the arc PA is a quad

ran t.

4 GRE AT AND SMALL C I RCLE S.

8 . Th e an g le subt e n d ed at th e ce n tr e of a sp h e r e by th e are ofa g r e a t circle wh ich joi n s th e p ole s of two g r e a t circle s is equal to th ei n clin ation of th e p lan e s of th e g r e at circle s.

Le t 0 be the cen tre of the Sphere, CD, OE t h e g reat circles int e rse ct in g at 0,

A an d B t h e poles of CD an d CE respectively.

Draw a g reat circle throug h A an d B,meetin g OD an d 0E at

M an d N respect ively. Then AO is perpen dicular t o 00,which is

a straig ht lin e in t h e plan e OOD 3 an d B0 is perpen dicular to 00,which is a straig ht lin e in t h e plan e 00E 5 therefore 00 is pe rpe ndicul ar to the plan e AOB (E uclid, X I . 4) an d therefore 00 is

perpen dicular to the st raig ht lin es OM an d ON,whi ch ar e in the

plan e AOB. Hen ceMON is t h e an g le of in clin ation of t h e plan esOOD an d O0E

’

. An d t h e an g le

AOB = AOM — BOM = BON BOM =MON.

9 . By the an g le between two g reat circles is mean t th e an g le

of i n clin a tion of th e p la n es of th e cir cle s . Thus,in t h e fig ure of

t h e precedin g Ar ticle, the an g le between the g reat circles CD an d

OE is the an g le MON.

I n the fig ure to Art . 6,sin ce P0 is perpen dicular to the plan e

AOB,every plan e wh i ch con t ain s P0 is at rig ht an g les t o the

plan e AOB. Hen ce the an g le between t h e plan e of an y Circlean d t h e plan e of a great circle which passes through its poles is

a rig ht an g le.

GREAT AND SMALL CIRCLES . D

1 0. Two g r e a t circle s bise ct e ach oth e r .

For Sin ce the plan e of each g reat circle passes th roug h the

cen tre of the sphere, the lin e of i ntersection of these plan es is a

diameter of the sphere, an d therefore also a diameter of each g reat

cir cle ; therefore the g reat cir cles ar e bisected at the poin ts wher e

they meet.

1 1 . [f th e arcs of g r e a t circle s join in g a p oin t P on th e surface

of a sp h e r e with two oth e r p oin ts A an d C on th e surface of th e

sp h e r e , wh ich ar e n o t a t Opp osite extr emi ti e s of a diame te r,be e ach of

th em equal to a quadran t , P is a p ole of th e g r e at circle th roug h

A an d C. (Se e the fig ur e of Ar t .

For suppose PA an d P0 to be quadran ts,an d O t h e cen tre of

t h e Sphere th en sin ce PA an d P0 ar e quadr an ts,the an g les P00

an d POA ar e r ig ht an g les . Hen ce P0 is at r ig ht an g les to t h e

plan e A00, an d P is a pole of the g reat cir cle A0.

1 2 . Great circles which pass th roug h t h e poles of a g reat

circle ar e called se con dar ie s t o that cir cle . Thus,in t h e fig ure of

Al t . 8 the poin t 0 is a pole of ABMN ,an d therefor e 0M an d 0N

ar e parts of secon daries to ABMN. A n d th e an g le between 0M

an d ON is measured by MN ; that is, th e an g le be twe e n an y two

g r e at cir cle s is me asur ed by th e ar e th ey in te rcep t on th e g r e a t circle

to wh ich th ey ar e se con dari e s .

1 3. I f from a p oin t on th e surface of a sp h e r e th e r e can be

drawn two arcs of g r e a t circle s, n ot p ar ts of th e same g r e a t circle ,th e p lan e s of wh ich ar e a t rig h t an g le s to th e p lan e of a g ive n circle ,th a t p oin t is a p ole of th e g ive n cir cle .

For,Sin ce t h e plan es of these arcs ar e at rig h t an g les to t h e

plan e of the g iven circle,t h e lin e in wh ich th ey in tersect is pe r

pe n di cular t o the plan e of the g iven circle, an d is th erefore t h eaxis of the g iven circle hen ce t h e poin t from wh ich t h e arcs ar edrawn is a pole of t h e circle.

6 SPHE RI CAL TRIANGLES.

1 4. To comp o/re th e ar e of a small circle subte n din g an y an g le

a t th e ce n tr e of th e circle with th e arc of a g r e a t circle subte n din g

th e same an g le a t i ts ce n tr e .

Le t ab be t h e arc of a small circle, 0 th e cen tre of the circle,

P the pole of the cir cle,0 the cen tre of the Sphere . Th roug h P

draw the g reat circles PaA an d PbB,meetin g the g reat circle

of wh i ch P is a pole,at A an d B respectively ; dr aw 0a, 0b, 0A,

OB. Then 0a, 0b, 0A,0B ar e all perpen dicular to OP,

because

the plan es a0b an d AOB ar e perpen dicular to OP ; therefore 0ais parallel to 0A,

an d Ob is parallel to 0B. Therefore th e an gle

aOb the an g le AOB (E uclid, X I . Hen ce,

ar c ab arc AB

radius 0d radius OA

th erefore, 33 g—g sin POa .

(Pla n e Trig on ome try, Art .

II. SPHE RICAL TRIANGLE S.

1 5 . Sph erical Trig on ometry in vestig ates the relation s which

subsist between the an g les of t h e plan e faces wh i ch form a solid

an g le an d the an gles at which the plan e faces ar e in clin ed to each

oth er .

SPHERICAL TRI ANGLE S . 7

1 6. Suppose that the an g ular poin t of a solid an g le is made

t h e cen tre of a sph ere then the plan es wh i ch form t h e solid an g le

wil l cut the sph ere in arcs of g reat cir cles. Th us a fig ur e will be

form ed on the surface of the sphere which is called a sph e rical

trian g le if it is boun ded by th r e e arcs of g reat circles this will be

the case when the solid an g le is form ed by the meetin g of thr e eplan e an g les. If the solid an g le be formed by the meetin g of

mor e th an th r e e plan e an gles, the correspon din g fig ur e o n thesurface of the sphere is boun ded by more than thr ee arcs of greatcircles

,an d is cal led a sp h er ical p olyg on .

1 7 . Th e th ree arcs of g reat circles wh ich form a spher ical

tr ian g le ar e called the sid e s of the Spher ical tr ian g le ; the an g les

formed by the arcs at the poin ts where they meet are called t h ean g le s of the spherical tr ian g le. (Se e Art .

1 8 . Thus, let 0 be the cen tre of a sphere,an d suppose a solid

an g le formed at O by the meetin g of three plan e an g les. Le t

AB,B0, 0A be t h e arcs of g reat circles i n wh ich t h e plan es cut

t h e sphere ; then ABC is a spherical tr ian g le, an d the arcs AB,

B0, 0A ar e its sides. Suppose Ab t h e tan g en t at A to the arc

AB,an d Ac the tan g en t at A to the arc AO

,the tan g en ts bein g

drawn from A towards B an d 0 respectively then the an g le bAcis on e of the an g les of t h e spherical trian g le . Sim ilarly an g les

formed in like man n er at B an d 0 ar e the other an g les of the

Spherical trian g le.

8 S PHERICAL TRIANGLES .

1 9 . Th e prin cipal part of a treatise on Spherical Trig on omet rycon sists of theorems r e lat in g to spherical trian g les it is thereforen ecessary to obtain an accurate con ception of a spherical trian g lean d it s part s.

It will be seen that what ar e cal led side s of a sphericaltrian g le ar e really ar cs of g reat circles

,an d these arcs ar e pro

por t ion al to t h e three plan e an g les which form t h e solid an g le

correspon din g t o the spherical t rian g le . Thus,in t h e fig ur e of

the precedin g Ar t icle,t h e arc AB forms on e side of t h e spherical

trian g le ABO, an d t h e plan e an g le AOB is measured by the fracarc AB

an d thus the arc AB is Proportion al to the an g le

AOB so lon g as we keep to the same Sphere.

Th e an g le s of a spherical trian g le ar e t h e in clin ation s of the .

plan e faces wh ich form the solid an g le ; for sin ce Ab an d Ac ar e

both perpe n dicular to 0A,t h e an g le bAc is t h e an g le of in cli n ation

of t h e plan es OAB an d OAO.

20. Th e letters A,B

,0 ar e g en erally used to den ote the

an g les of a spherical tri an g le, an d the letters a,b,c are used t o

den ote the side s. A S in t h e case of plan e trian g les, A,B

,an d 0

may be used to den ote the n umerical values of t h e an g les expressed

in t e rms of an y un i t, provided we un derstan d di stin ctly what the

un it is. Thus,if the an g le 0 be a rig ht an g le, we may say that

W

0 or th at 05 ,

accordin g as we adopt for the un i t a de

g ree or t h e an g le subt en ded at t h e cen tre by an are equal to t h eradius . So also

,as t h e sides of a spherical trian g le ar e propor

t ion al to the an g les subten ded at t h e cen tre of the sphere, we

may use a, b, c to den ote the n umerical values of those an g les

in t erms of an y un it . We shall usually suppose both the an g les

an d Sides of a Spherical trian g le expressed in circu lar me asur e .

(Plan e Trig on ome try, Art .

SPHERICAL TRIANGLES. 9

2 1 . I n future, un l ess t h e con trary be distin ctly stated,an y

ar e drawn on the surface of a Sph ere will be supposed to be an arc

of a g r e at cir cle.

2 2 . I n spherical trian g les each side is restr icted to be lessthan a semicir cle this is of course a con ve n tion

,an d it is adopted

because it is foun d con ven i en t.

Th us,in t h e t h e arc ADE B is g reater than a semicir

cumfe r e n ce,an d we mig ht, if we pleased, con sider ADE B, AO,

an d B0 as formin g a trian g le, havin g it s an g ular poin ts at A,B,

an d 0 . But we ag ree to exclude such trian g les from our con

Sideration ; an d the tr ian g le havin g it s an g ular poin ts at A,B,

an d 0, will be un derstood to be th at formed by AFB,B0 an d 0A.

2 3 . From t h e restri ction of the precedin g Article it will

follow th at an y an g le of a sp h e rical trian g le is le ss tha n two rig h t

an g le s.

For suppose a trian g le formed by B0, 0A,an d BE DA,

h avin gthe an g le BOA greater th an two rig h t an gles. The n suppose Dto den ote the poin t at which the arc B0, if produced, will meet

AE then BBB is a semi circle by Art . 10,an d therefore BE A

is g reater than a semicir cle t h is n ot on e

of those whi ch we con sider.

( 1 0 )

III. SPHE RICAL GE OME TRY .

2 4. Th e relation s between t h e Sides an d an g les of a Spheri cal

Trian g le, which are in vestig ated in t reatises on Spherical Trig on ometry

,ar e chiefly such as in volve the Tr ig on ome trical Fun ction s

of the sides an d an g les. Before proceedin g t o th ese,‘

however,we

Shall collect,un der the head of Spherical Geometry, some theorems

which in volve the sides an d an gles th emse lve s, an d n ot their trig o

n om e t rical ratios.

2 5 . Polar Man g le . Le t ABO be an y sph erical trian g le, an dlet the poin ts A’

,B’

, 0’

be those poles of the arcs B0, 0A,AB

respectively which lie on t h e sam e Sides of them as the Opposite

an g les A,B

,0 then the trian g le A

'

B’

0'

is said to be the p olar

tr ia n g le of t h e trian g le ABO.

Sin ce there ar e two poles for each side of a Spherical trian g le,e ig h t trian g les can be formed havin g for their an g ular poin ts polesof t h e sides of t h e g iven trian g le but there is on ly on e trian g le in

which these poles A'

,B

'

,0

’ lie towards the same parts with thecorrespon din g an g le s A,

B, 0 an d thi s is the trian g le which is

kn own un der t h e n ame of the p olar trian g le .

Th e trian g le ABO is called the p rimitive trian gle with respectt o the trian gle

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1 2 SPHERICAL GEOMETRY.

supplemen t of the an g le A. Th i s we may express for Shortn essthus ; B

’

0’

is the supplemen t of A. Similar ly it may be shewnthat 0 ’

A’

is the supplemen t of B, an d A’

B’

t h e supplemen t of 01

A n d Sin ce ABO is the polar trian g le of it follows thatB0

,0A,

AB ar e respectively the supplemen ts of A’

,B’

, 0’ that is

,

A’

,B’

, 0’

ar e respect ively the supplemen ts of B0, 0A,AB.

From these properties a prim itive trian g le an d it s polar t rian g le ar e sometimes called supp leme n tal trian g le s .

Thus,if A

,B

,0,a, b, 0 den ote respe ctively the an g les an d

the Sides of a Spherical tri an g le, all expressed in circular measure,an d A

’

,B’

,a’

, b’

,0’ those of t h e polar trian g le, we have

A'= 7 r — a

,B

'z

‘lr — b

, 0'= 7 r - 6 ,

2 8 . Th e precedin g result is of g reat importan ce ; for if an y

g en eral th eorem be demon strated with respect to the sides an d thean g les of an y Spherical trian g le it holds of course for the polar

trian g le also. Th us an y such th e or em wi ll r emain tru e wh e n th e

an g le s ar e chan g e d in to th e suppleme n ts of th e corr e sp on din g side s

an d th e side s in to th e supp leme n ts of th e corr esp on din g an g le s . We

shall se e several examples of th i s pri n ciple in the n ext Chapter .

2 9 . An y two side s of a sp h e r ical tria n g le ar e tog e th e r g r e a te r

tha n th e th ird side . (S e e the fig ure of Art .

For an y two of the three plan e an g les wh ich form the solid

an g le at 0 ar e tog ether g reater than the th i rd (E uclid, XI.Therefore an y two of t h e arcs AB

,B0, CA,

ar e tog ether g reater

From this proposition it is obvious that an y side of a Spherical

trian g le is g reate r than the differen ce of the oth er two .

30. Th e sum of th e th r e e side s of a sp h e rical trian g le is le ss th an

th e circumfer e n ce of a g r ea t circle . (Se e the figur e of Ar t .

SPHERICAL GEOMETRY. 1 3

For the sum of the three plan e an g les which form the solid

an gle at O is less than four r ig ht an g les (E uclid, XI . 2 1) therefore

AB BC 95—4 is less than 2 7r,

0A 0A 0A

therefore,

AB B0 0A is less than 2 7 : x 0A

that is,the sum of the arcs is less than the cir cumferen ce of a

g reat circle .

3 1 . Th e proposition s con tain ed in t h e precedin g two Ar t iclesmay be ext en ded. Thus

,if there be an y polyg on which h as each

of it s an g les less t h an two r ig ht an g les, an y on e sid e is le ss th an th e

sum of a ll th e oth e rs . Thi s may be proved by repeated use ofA r t . 2 9 . Suppose, for example, that the fig ure h as four sides

,an d

let the an g ular poin ts be den oted by A,B

,0,D . Then

AB +B0 is g reater than A0 ;

therefore, AB B0 0D is g reater than A0 0D,

an d afor tior i g r eater than AD .

Ag ain ,if ther e be an y polyg on which h as each of it s an g les

less than two r ig ht an g les, th e sum of i ts side s wi ll be le ss th an th e

cir cumf e r e n ce of a g r e a t cir cle . This follows from E uclid,XI . 2 1

,

i n t h e man n er shewn in Ar t . 30.

32 . Th e th r e e an g le s of a sp h e rical trian g le ar e tog e th e r g r eate r

th an two r ig h t an g le s an d le ss th an six r ig h t a n g le s .

Le t A,B

, 0 be the armle s of a sph erical trian g le 5 le t a’

,b'

,c’

be t h e side s of the polar tr ian g le. Then by Ar t . 30,

a’

b'

c’

is less th an 2 7r,

t h at is, w —A + 7r —D + 7 r — 0 is le ss t h an 2 7 1-5

therefore,

A B 0 is g reater than 7 r.

An d sin ce each of the an gles A,B, 0 is less th an W

,t h e sum

A B 0 is less than 3m

1 4 SPHERICAL GEOMETRY.

33. Th e an g le s at th e base of an isosce le s sp h e r icalequal.

Le t ABO be a Sph erical trian g le havin g A0 B0 ; let 0 be t h ecen tre of the sphere. Draw tan g en ts at t h e poin ts A an d B to thearcs A0 an d B0 respectively 5 these will meet 00 produced at thesame poin t S, an d AS will be equal to BS .

Draw tan g en ts AT,BT at t h e poin ts A,

B to the arc AB 3 thenAT z TB join TS . I n the two trian g les SAT, SBT t h e sidesSA

,AT,

TS ar e equal to SB,BT

,TS respectively ; therefore t h e

an g le SAT is equal to the an g le SBT 5 an d these ar e the an g les atthe base of t h e spherical t rian g le.

Th e fig ure supposes A0 an d B0 to be less th an quadr an ts if

th ey ar e g reater than quadran ts t h e tan g en t s to A0 an d B0 will

meet on 00 produced throug h 0 in stead of t hroug h 0,an d the

demon stration may be completed as before. If A0 an d B0 ar e

quadr an ts, the an g les at the base ar e r ig ht an g les by Ar ts. 1 1

an d 9.

34. If two an g le s of a sp h e r ical tr ian g le ar e equal, th e opp osite

sid e s ar e equa l.Sin ce t h e primitive trian g le h as two equal an g les, the polar

trian g le h as two equal sides 3,

therefore in t h e polar t rian g le t h e

an g les opposite the equal sides ar e equal by Art . 33 . Hen ce in

t h e prim it ive trian g le the sides opposite the equal an g les ar e

equal.

SPHERICAL GEOMETRY. 1 5

35 . If on e an g le of a sp h e rical tr ian g le be g r eate r th an an o

th e r, th e sid e opp osi te th e g r e a te r an g le is g r e ate r th e sid e

opp osite th e le ss an g le .

Le t ABO be a spher ical trian g le, an d le t t h e an g le ABO be

g reater than the an g le BAO : then the side AO will be g reaterthan the side B0 . A t B make the an g le ABD equal to the an g leBAD then BD is equal to AD (Art . an d BD D0 is g reater

than B0 (Art . 2 9)3 therefore AD +D0 is greater th an B0 3 that

is,A0 is greater than B0 .

36 . If on e side of a sp h erical trian g le be g r e at e r th an an oth e r,

th e an g le opp osite th e g r e ater side is g r e a t e r th an th e an g le opp osi t e

This follows from the precedin g Article by mean s of the polar

trian g le.

Or thus ; suppose t h e side A0 g reater than the side B0,then

the an g le ABO will be g reat er than the an g le BAO. For the

an g le ABO can n ot be less than the an g le BAO by Ar t . 3 5,an d

the an g le ABO can n ot be equal to the an g le BAO by Ar t . 34 5

therefore the an g le ABOmust be g reater than the an g le BA0.

This Chapter mig ht be exten ded 5 but it is un n ecessary to do

so because t h e Trig on ometr ical formulae of the n ex t Chapt er sup

ply an easy method of in vest ig atin g t h e theorems of Spherical

Geometry. Se e Arts. 5 6, 5 7 , an d 5 8 .

IV. RE LATIONS BE TWE E N THE TRIGONOME TRICALFUNCTIONS OF THE SIDE S AND THE ANGLE S

OF A SPHE RICAL TRIANGLE .

37 . To exp r e ss th e cosin e of an a n g le of a trian g le i n te rms ofsin e s an d cosin e s of th e sid e s .

Le t ABC be a spherical trian g le, 0 the cen tre of t h e Sph ere .Le t th e tan g en t at A to t h e arc AO meet 00 produced at E

,an d

let the tan g en t at A to th e arc AB meet 0B produced at D ; j oinE D. Thus the an g le E AD is t h e an g leA of the spherical trian g le,an d the an g le E OD measures the side a.

From the trian g les ADE an d 0DE we have

DE 2 = AD2+ AE

S2AD AE cos A

,

DE”: 0D

2+ 0E

2Q OD . 0E cos a ;

also the an g les OAD an d 0AE ar e rig ht an g les, so that

0D9= 0A

2+ AD2

an d Hen ce by subtraction

we have

20A2+ 2AD . AE cos A— 2 0D . 0E cos a

,

0A 0A AE AD

oe o zfioa on

that is cos a cos 6 cos c sin 6 sin 0 cosA.

therefore cos a cosA 5

cos 0

sin 0sin c

RELATIONS BETWEEN THE FUNCTIONS . 1 7

38. We h ave supposed, in the con struction of the precedin gAr ticle

,th at t h e Sides wh i ch con tain the an g le A ar e less than

quadran ts,for we h ave assumed that the tan g en ts at A meet 0B

an d 00 r espectively produced. We must n ow shew that theformula obtain ed is true when these sides ar e n ot less th an quad

ran ts . Th is we shall do by Special examin ation of the cases inwh ich on e side or each side is g reater than a quadran t or equal to

(1) Suppose on ly o n e of t h e sides wh i ch con tain t h e an g le A

t o be g reater th an a quadran t,for example

,AB. Produce BA

an d B0 to meet at B’

; an d put AB’

c'

, 0B’

a'

.

Th en we h ave from t h e trian gle AB’

C,by wh at h as been

already proved,

cos a’: cos bcos c

'

+ Sia in c'

cos BAG5

but a'= 7r

— a,c'= 7r — c

,B

'

A0 = 7 r —A,

‘ thus

cos a = cos bcos c + sin b sin c cos A.

(2) Suppose both t h e sides wh ich con tain t h e an g le A to be

g reater than quadran ts. Produce AB an d A0 to meet at A’ put

A’

B c'

,A

’

O b’

; then from the trian g le A’

B0, as before,

cos a z z cos b ’ cos c’

+ sin 6’sin c’ cosA’

;

but b'z w - b

,c’: 7r — c

,A

'z A ; th us

T. S. T.

1 8 RELATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS

(3) Suppose that on e of the Sides wh ich con tain the an g le Ais a quadran t

,for example

,AB on A0, produced n ecessary;

AD equal t o a quadran t an d draw BD. If BD is a quadran t

B is a pole of A0 (Ar t in th is case a an d A z z-z

r

as w e ll

as 0 Thus t h e formula to be verified reduces t o th e iden tity

0 0. If BD be n ot a quadran t,t h e tr ian g le BDO g ives

cos a. cos CD cos BD Sin CD Sin BD cos 0DB,

an d cos 0DB 0,

cos CD cos Sin 6,

cos BD cos A

cos a = sin bcosA ;

an d th is is wh at t h e formula i n Ar t . 3 7 becomes wh en 0 g(4) Suppose th at both t h e sides wh ich con tain t h e an g le A

ar e quadran ts. Th e formula th en becomes cos a = cosA an d th isis obviously true

,for A is n ow the pole of B0, an d thus A a .

Th us t h e formula in Art . 3 7 is proved to be un iversally true.

39 . Th e formula in Art . 37 may be applied to express thecosin e of an y an g le of a trian g le in terms of Sin es an d cosin es of

the Sides 5 thus we h ave the th ree formulae ,cos a = cos bcos c + Sin bsin c cos A

,

cos b= cos c cos a + sin c Sin a cosB,

cos c.= cos a cos b+ sin a sin bcos 0.


Le t ABO b e a sph erical trian g le, 0 the cen tre of t h e sph ere .Take an y poin t P in 0A,

di aw PD perpen dicular to t h e plan e

B00,an d from D draw DE ; DE perpen dicul ar to 0B, 00 r e spe c

t iv e ly j oin PE , PE , 0D.

Sin ce PD is perpen di cular to t h e plan e B00, it makes rig ht

an g les with every straig ht lin e meetin g it in that plan e h en ce

PE 2 = PD”+DE

2 =P020D

’+DE 2 =P0

2OE

”

;

thus PE 0 is a right an g le. ThereforePE OP Sin POE OP Sin c ;an d PD =PE Sin PE D =PE Sin B = 0 ? Sin c sin B.

Similarly,PD OP Sin 6 Sin 0 ; therefore

OP Sin c Sin B ==0P sin bsin 0 ;

sin B Sin 6

Sin 0 8 11 1 c

Th e fig ure supposes b, c, B,an d 0 each less than a rig h t an g le ;

it wi ll be foun d on examin ation that the proof will h old when th e

fig ure is modified to meet an y case which can occur . If, for

in stan ce,B alon e is g reater than a rig ht an g le, the poin t D will

fall beyon d 0B in stead of between 0B an d 00 ; then BE D will

be th e suppleme n t of B, an d thus sin BE D is still equal to Sin B.

43 . To sh ew th a t cot a sin b cot A sin 0 cos 6 cos 0.

We h ave cos a cos 6 cos c sin 6 sin 0 cosA,

cos c = cos a cos b+ sin a sin b cos 0,

Sin 0sm c z sm a

S I n A

t h erefore

OF THE SIDES AND THE GLE. 2 1

Substitute t h e values of cos c an d sm c equation

th ussin a sin b cos A sin 0

cos aS i n A

by t ran sposition

cos a sin’b= Sin a sin bcos bcos 0 + sin a sin bcot A sin 0 ;

divide by sin a Sin b th us

cot a sin b z cos bcos 0 + cot A sin 0.

44. By in te rchan g in g t h e letters five other formulae may be

obtain ed like that in t h e precedin g Article the whole six formulaewill be as follows

cot a sin b z cot A Sin 0 + cOS bcos 0,

cot bsin a z cot B Si n 0 + cos a cos 0,

cot 6 sin c = cot B sin A+ 008 0 cos A,

cot c sin b cot 0 sin A+ cos b cosA,

cot e sin a = cot 0 sin B + cos a cosB,

cot a sin c=cot A sin B + cos 0 cos B.

45 . To expr e ss th e sin e , cosin e , an d tan g e n t, of half an an g le

of a trian g le asfun ction s of th e side s.

cos a — cos bcos c.

Sin bsin cWe h ave

,by Art . 37 , cosA

cos a — cos bcos c cos b— c — cos ath erefore I cosA 1

lSin b sin c sin b sin c

th erefore 4 8m (a b C) i (a b Cl2 sm bsm c

Le t 2 s = a + b+ c, so th at s is h alf t h e sum of t h e sides

t h e trian gle ; th en

a + b _ c = 2 s — 2 c= 2 (s —c), a — b+ c = 2 s— 2b= 2 (8 — b)3

A Sin (s — b)sin (s — c)sin bsin c


cos a — cos b cos c‘

cos a — cos b+ c

A lso,1 + cosA z l +

Sin b sin c sin bsin c

therefore

% (a + h 0)Sin % (6 + c — a) Sin s sin (s — a)

sin bsin c Sin bsin c

sin 1) sin c

A AFrom t h e expressi on s for Si n

5an d cos

5we deduce

Sin (s — b)Sin (s — c)sin s si n (s — a)

Th e positive Sig n must be given to the radicals wh ich occur in

Athis Article, because is less th an a rig ht an gle, an d th ere fore its

Si n e,cosin e

,an d tan gen t ar e all positive.

46 . Sin ce Sin A 2 Sin ; cos we Obtain

Si n s sin (s a)sin (8 6)sin (8 c)}3

.

I t may be shewn that t h e expression for sin A in Art . 40

ag rees wi th the presen t expression by putt in g the'

n umerator of

that expression in factors, as in Plan e Tr ig on ome try, Art . 1 1 5 .

We Shall fin d it con ven ien t to use a symbol for the radical in the

value of sin A we shal l den ot e it by n,so that

Sin s sin (s — a)sin (s — b)sin (s — c),

4n21 cos’a cos

gb cos

’c 2 cos a cos 6 cos c.

OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE . 2 3

47 . To e xpr e ss th e cosin e of a sid e of a tr ian g le in te rms of

I n the formula of Ar t . 37 we may, by Art . 2 8,ch an g e the

sides in to the supplemen ts of the correspon din g an g les an d the

an g le in to the supplemen t of the correspon din g side thus

cos (7: —A)= cos (1r — B)cos (7: 0) sin (7 : -B)sin (7r —0)cos (1 : - a),

th at is, cos A —,cos B cos 0 + sin B sin 0 00s a .

Similarly cos B cos 0 cosA sin 0 sin A cos b,

cos 0 = — cosA cos B + sin A sin B cos c.

48. Th e formulae in Art . 44 wil l of course remain true wh en

t h e an g les an d Sides ar e chan g ed in to the supplemen ts of t h e cor

respon din g sides a n d an g les respectively ; it will be foun d, h ow

ever,that n o n ew formulae ar e thus obtain ed

,but on ly t h e same

formulae over ag ain . This con sideration will furn i sh some assist

an ce in retain in g th ose formulae accurately in the memory.

49 . To expr e ss th e sin e , cosin e , an d tan g e n t, of h alf a side of a

tria n g le asfun ction s of th e an g les .

cosA + cos B cos 0.

sin B sin 0We have

,by Art . 47 , cos a

th erefore

cos A+ cosB cos Cl — cos a z l

sin B sin 0 sin B sin 0

l

th ereforea B 0)cos 2 (B 0 A)2 sm B S I n 0

Le t t h e n B + 0—A = 2 (S —A), th erefore

cos S cos (S A)sin B sin 0

2 4 RE LATIONS BETWEEN THE TRIGONOMETRICAL FUNCTIONS

Also 1 + cos a = 1“ OMB — 0)

sin B sin 0 sin B sin 0

therefore

,_a cos % (A — 0) cos (S B)cos (S 0)

2 Sin B Sin 0 Sin B sin 0

a cos (S B)cos (S 0)2 sin B sin 0

a cos S cos S —AHen ce tan

2 cos (S — B)c(os (S30)

Th e positive Sign must be g iven t o the radicals wh ich occur in

th is Article, because 3is less th an a righ t an gle.

50. Th e e xpre ss mn s I n t h e precedin g Article may also be

Obtain ed immediately from th ose g iven in Art . 45 by mean s of

Art . 2 8.

It may be remarked th at th e values of sin 3, g, an d tan -gar e r eal. For S is greate r than on e rig h t an gle an d less than th reer ig h t an g les by Art . 32 th erefore cos S is n e g ative . An d in th e

polar trian g le an y Side is less than the sum of t h e other two th us7r —A is less than 7 r—B 7r 0 ; therefore B + 0—A is less th an

th erefore S A 1 8 less th an —

2an d B + 0 —A is algebraically

r

g reater th an — r, so that S —A is algebraically greate r th an

therefore cos (S — A) is p osit ive . Similarly also cos (S —B) an d03 a a

cos (S 0)ar e positive. Hen ce th e values of sin éan d tan

2

ar e real.

5 1 . Sin ce sin a 2 Sin6

5cos3, we obtain

sin sin 0 cos S oos (S A)cos (S B)cos (S

We sh all use N for cos S cos (S A)cos (S B)cos (S

OF THE SIDES AND THE ANGLES OF A SPHERICAL TRIANGLE. 2 5

5 2 . To demon strate Nap ie r’

s An alog ie s.

sin A sm B

sin a sin hWe have m suppose

th en,by a th eorem of Al gebra,

Sin A Sin Bm

S I n a sm b

an d alsoS i n a sm 6

Now cosA+ cosB cos 0 = sin B sin 0 cos a =m sin 0 sin bcos a,

an d cosB + cosA cos 0 = sin A sin 0 cos b=m sin 0 sin a cos 6,

therefore,by addition ,

th erefore by ( 1)we h ave

sin A + sin B sin a + sin 6 1 + cos 0

cosA + cosB Sin (a + b) sin 03

th at is,

tan B)

Similarly from (3)an d (2)we h ave

sin A — sin B sin a — sin h l + cos O

cos A + cos B sin (a + b) Sin 0

sin (a b) 0

sin 5 (a b) 2

3

th at is, tan 504 - B)

By writin g 7 r —A for a,an d so on in (4) an d (5)w e obtain

cos 4 (A B) ccos 2

tan % (a gTh e formulae (7) may be put in t h e form of pro

portion s or an alogies, an d are called from th eir discoverer Nap ies


An alog i es the last two may"

be demon strated without recurri n gt o t h e polar trian g le by startin g with the formulae in Art . 39.

5 3 . I n equation (4) of t h e precedin g Article,cos 4(a b) an d

0cot —

2are n ecessarily positive quan tities ; h en ce t h e equation

Shews that tan % (A +B) an d ar e of t h e same Sig n

thus (A B)an d (a b) ar e either both less than a rig ht an g le

or both g reater than a rig ht an g le. This is expressed by sayin g

that 5 (A B)an d 5 (a b) ar e of th e same af e ction .

5 4. To demon strate De lam br e ’s An alog ie s .

We h ave cos c = cos a cos b+ sin a sin b cos 0 ; th erefore

— sin fiC’

)

{ 1 + cos (a — b)}

therefore cos’z

5 c cos”

4(a 6)cos“

50 cos”

% (a + 6)Sin2

50.

Similar ly, Sin2

4c sin”

4(a 6)cos2

50 Sin2

4 (a 6)sin?

50.

Now add un ity to the square of each member of Napier ’s first

two an alog ies h en ce by t h e formulae just proved

cos21} c

sin g % osw im—B)

E xtract t h e square roots ; thus, Sin ce an d

are of th e same affection , we obtain

cos % (A +B)

cos % (A — B)sin §

Mul tiply the fir st t wo of Napier ’s an alogies respect ively bythese resul ts ; th us

— b)cos % 0

sin % (A— B) — b)cos % 0


this Shews that 4(A B) is positive, n eg ative, or zero, accordin g as

(a b) is positive, n eg ative, or z ero ; thus we obtain all the r esults in cluded in Arts.

5 7 . If two trian g le s h ave two sid es of th e on e equal to two

side s of th e oth e r,e ach to e ach

,an d likewise th e in clude d an g le s

wual, th e n th e ir oth e r an g le s wi ll be equal, e ach to e ach,an d like

wise th e ir base s will be equal.We may sh ew that the bases ar e equal by applyin g the first

formula i n Art . 39 to each trian g le, supposin g b, c, an d A the

same in the two trian g les ; then the remain i n g two formulae of

Ar t . 39 will sh ew that B an d 0 ar e the same in the two trian g les.

It should be observed that the two trian g les in th is case ar en ot n ecessar ily such th at on e may be made to coin cide wi th th e

oth e r by sup e rp osi tion . Th e Sides of on e may be equal to those of

t h e oth e r,each to e ach , but in a re verse as in t h e followin g

figure s.

Two tri an g le s wh i ch ar e equal in th is man n er ar e sai d to be

symme trically equal wh en th ey are equal so as to admit of super

posit ion th ey are said to be absolu te ly equal.

5 8. If two sp h erical trian g le s h ave two side s of th e on e equal totwo side s of th e othe r

,e ach to each

, bu t th e an g le wh ich is con tain ed

by th e two side sqf th e on e g r e ate r than th e an g le wh ich is con tain e d

by th e two side s wh ich ar e equal to th em of th e oth e r, th e base of th at

OF THE SIDES AND THE ANGLES OF A. SPHERICAL TRIANGLE. 2 9

wh ich h as th e g rea te r an g le wi ll be g r e a t e r th an th e ba se of th e

oth e r ; an d con ve rse ly.

Le t b an d c den ote t h e sides wh ich ar e equal in t h e t wo t ri

an g les le t a be the base an d A the opposite an g le of on e trian gle,an d a

’

an d A’

similar quan tities for the other . Then

cos a = cos bcos c+sin b Sin c cosA,

th erefore cos a cos a’

sin 6 sin c (cos -A cos A’

sin % (a - a’

)= sin bsin c sin § (A+A’

)sin 11; (A

th is Shews tha t a‘)an d (A A

'

)ar e of t h e same Sign .

5 9. If on a sp h e r e an y poin t be take n with in a circle wh ich

is n ot i ts p ole , of a ll th e arcs wh ich can be drawn from th a t p oi n t

to th e circumf e r e n ce , th e g r e atest is th a t i n wh ich th e p ole is , an d th e

o th e r par t of that p roduce d is th e le ast ; an d of an y oth e rs,th a t wh ich

is n e are r to th e g r e a test is a lways g r e a t e r th an on e mor e r emote ; an d

from th e same p oin t to th e circumf e r e n ce th e r e can be drawn on ly

two (was wh ich are equal to e ach o th e r,an d th e s e make equal an g les

with th e sh or te st ar e on opp osit e side s of i t. .

Th is follows readily from t h e precedin g th ree Articles.

60. We will g ive an oth er proof of the fun damen tal formulaein Art .

’

39,wh ich is very Simple

,requirin g on ly a kn owledg e of

the elemen ts of Co-ordin ate Geometry.

Suppose ABO an y sph erical trian g le, 0 t h e cen tre of t h e

Sph ere, take 0 as t h e orig ino

of ca ordin at e s,an d le t the axis of z

pass throug h 0 . L e t £01 , y ,, z

,be the co -ordin ates of A

,an d x

2 ,

y2 , z , those of B ; let r be t h e radius of t h e sphere. Th en’

thesquare on the straig ht lin e AB is equal t o

(” I way (!l/x "

ye l+ (z 1 zslz

’

an d also to r”

r‘32 r

icos AOB

30 EXAMPLES .

an d as,

”

y,2

z,

2r’

,x,

”

y,2

z f r”

,th us

also, y,y, z

lz,

r”cosA0B.

Now make the usual substitution s I n passin g from rectan gular

to po lar co—ordin ates, n amely,

zlz r cos 0 a

l= r sin 0

,cos cpl , y l

z r sin Olsin qt

z,= r cos 0

2 ,a,= r sin 0

,cos 952 , g g

= r sin 6,sin (fig ;

thus we obtain

cos 9, cos 0, sin 0,Sin 9

,cos (qt l qt g) cosAOB,

t h at is, in th e ordin ary n otation of Sph erical Trig on ometry,

cos a cos b+ s1n a sm b cos 0 = cos c.

Th is meth od h as the advan tage of g ivin g a p e rfe ctly g e n e ral

proof, as all the equation s used ar e un iversally.

true.

E XAMPLES .

1 . If A a,sh ew th at B an d b are e qual or supplemen tal, as

also 0 an d c.

2 . If on e an g le of a tr ian g le be equal t o t h e sum of t h e other

two,the g reatest Side is double of the distan ce of its mi ddle poi n t

from the opposite an gle.

3 . Wh en does the polar trian g le coin cide with t h e primittrian g le ?

4. If D be the mi ddle poin t of AB,sh ew th at

cos AO cos B0 2 cos 4AB cos 0D.

5 . If two an g les of a sph erical,

trian g le be respectively e qual

to t h e sides opposite t o them,shew that t h e remain in g side is the

supplemen t of the remain in g an g le or else th at the trian g le h as

two quadran ts an d tworig ht an gles, an d then t h e remain i n g sideis equal to the remain in g an g le.

EXAMPLES. 3 1

a A6 . I n an equilateral trian gle, shew that 2 cos sin

2

2_

hen ce deduce the limits between which the sides an d t h e an g les of

an equilateral trian g le ar e restricted.

7 . I n an equilateral trian gle, sh ew th at tan 2 1 2 cosA ;

8 . I n an equilateral trian g le, sh ew th at se cA 1 se c a.

9. If t h e th ree sides of a sph erical trian g le be h alved an d

a n ew trian gle formed,t h e an g le 0 between t h e n ew sides

4

6an d g

is g iven by cos 0 cos A 4t an 2tan 5Sin20.

10. AB, CD ar e quadran ts on t h e surface of a Sph ere in ter

se ct in g at E ,th e extremities bein g join ed by great cir cles : sh ew

that

cosAE 0 cosAO cos BD cos B0 cos AD.

1 1 . If 6 c = 7r,sh ew th at sin 2B Sin

1 2 . If DE be an arc of a great circle bisectin g t h e sides AB,

A0 of a sph erical trian g le at D an d E,P a pole of DE

,an d PB,

PD,PE

,P0 be j oin ed by arcs of great circles

,sh ew th at t h e an gle

BPO= twice t h e an g le DPE .

1 3 . I n a sph erical trian gle sh ew th at

sin bsin c + cos bcos c cosA= sin B sin 0 — cosB cos 0 00s a .

1 4. If D be an y poin t in t h e side B0 of a trian g le, sh ew th at

cosAD sin B0 cosAB sin D0 cosAO sin BD .

1 5 . I n a sph erical trian g le sh ew that if 9, (I), 30be the len gth s

of arcs of g reat circles drawn from A,B, 0 pe rpen dicular to t h e

opposite Sides,sm a sin 0= sin b Sin gs 8111 6 8111 l

J (1 — cos’a - cos

2 b — cos"c + 2 cos a cosb cos c).

32 SOLUTION OF RIGHT-ANGLED TRIANGLES.

1 6. I n a spherical trian g le, if 9, ct , Ill be the arcs bisectin g t h e

an g les A,B, 0 respectively arid termin ated by the oppOSit e sides

,

shew that

cot a + cot b+ cot c.2 2 2

1 7 . Two ports are in t h e same parallel of latitude, th eir com

mon latitude bein g l an d their differen ce of lon g itude 2A: shew

that the savin g of distan ce in sailin g from on e t o t h e other on the

g reat circle, in stead of sailin g due E ast or West , is

2 r {Acos l sin'" 1

(sin Acos

Abein g expressed in circular measure, an d r bein g t h e radius of

t h e E arth .

1 8 . If a sh ip be proceedin g un iformly alon g a g reat circle an d

t h e observed lat itudes be Znl,,la,

at equal in tervals of time,in

each of wh ich t h e distan ce traversed is s, shew that

- 1Sin 2 (ll l

a)COS (ll la)

s ==r cos8 111 l

,

3

r den otin g t h e E arth’s radius : an d shew th at the ch an g e of lon gi

t ude may also be foun d in terms of the th ree latitudes.

V . SOLUTION OF RIGHT-ANGLE D TRIANGLE S.

6 1 . I n every Sph erical trian g le th ere ar e six elemen ts, n amely,t h e three sides an d t h e three an g les, besides t h e radius of the

Sphere,wh i ch is supposed con stan t. Th e solution of spherical t ri

an g les is the process by wh i ch, when the values of a sufficien tn umber Of the six elemen t s ar e g iven , we calculate t h e values of

the remain i n g elemen ts. I t will appear, as we proceed, th at w hent h e values of three of the elemen ts ar e g iven , those of the remainin g three can g en erally be foun d. We beg in with the rig ht-an g ledt rian g le here two elemen ts, in addition to t h e rig ht an g le, will be

supposed kn own .

SOLUTION OF RIGHT-ANGLED TRIANGLES. 33

62 . Th e formulae r equisite for the solution of rig h t-an g led

trian g les may be obtain ed from the precedin g Chapter by sup

posin g : on e of the an g les a an g le, as 0 for example. They

may also be obtain ed very easily in an in depen den t man n er,as

we will n ow shew.

Le t ABO be a sph erical h avin g a rig h t at 0 3

le t 0 be t h e cen tre of the sphere. From an y poin t P in OA draw

PM perpen dicular to 00,an d from M draw MN perpen dicular to

0B,an d j oin PN. Then PM is perpen di cular to MN

,because the

plan e AOU is perpen dicular to the plan e B00 ; hen ce

PN2 = PM 2

+M7 2 = 0P

20M 2

+ 0M 20N

2 = GP2ON

”

,

th erefore PNO is a rig h t an gle. An d

ON ON 0Mt h t is cos c OS 6

OP UM.

OP,

8 ;3

- c “ COS

P211 PM PN

OP PN'

0P,t h at 1 s

,sm b= smB sm c

Sim ilar ly sin a = sin A sin c

MN MN PN

ON PN’

0 V,t h at is

,t an a = cot an c

J.

Similarly tan b cosAtan 0

0M MN.

0M ,th at is

,t an bz tan Bsin a

S imilarly tan a t an Asin b

T. S. T.

3 4 S OLUTION OF RIGHT-A NGLED TRIANGLES.

Multiply t og ether the two formulae (4) t hus,

tan A tan B —tSn a tim b 1 1

8 1 11 a S i n b cos a cos b cos c

there fore cos c cotA cotB . .

Multiply crosswise the secon d formula in (2) an d t h e fir stin (3) t h us Sin a cot an c z tan a sin A sin c ;

therefore cos B Sin A cos b by

cos B = Sin A cos b

cos A ==sin B cos a

These Six formulae comprise t e n equation s ; an d thus we can

solve every case of r ig ht -an g led trian g les. For every o n e of these

t e n equation s is a dist in ct combin ation in volvin g three out of the

fivequan tities a , b, c, A,B ; an d out of five quan tities on ly t e n

combin ation s of three can be formed. Thus an y two of the fivequan tities bein g g iven an d a third required

,some on e of the pr e

cedin g te n equation s will serve to determ in e th at thirdquan tity.

i

63 . AS we have stated,the above six formulae may be Oh

t ain e d from those g iven in the precedin g Chapter by supposin g 0 a

r ig ht an g le. Thus ( 1) follows from Ar t . 39, (2) from Art . 4 1 ,

(3) from the fourth an d fifth equation s of Art . 44, (4) from the

first an d secon d equation s of Ar t . (5)from the t hird equationof Ar t . 47 (6)from the first an d secon d equation s of Art . 47 .

Sin ce the six formulae may be obtain ed from those g iven in

the precedi n g Chapter wh i ch have been proved to be un iversally

true, we do n ot stop to shew that the demon stration of A rt . 62

may be applied to every case whi ch can occur ; t h e studen t may

for exercise in vestig ate the modification s which will be n ecessary

when we suppose on e or more of thequan tit ies a, b, c, A,B equal

to a rig ht an gle or g reater th an a righ t an g le.


Th e rig h t an g le is left"

out of con sideration ; t h e two.

sideswh ich in clude the r ig h t an g le, the complemen t of the hypoten use,an d the complemen ts of the other an g les ar e called the circular

parts of th e trian gle. Th us th ere ar e five circular parts, n amely,7 1

'

A7 7

2 2

roun d a circle in t h e order in wh ich th ey n aturally occur wi threspect to the trian g le.

An y on e of the five par ts may be selected an d call ed t h e

middle p ar t , th en the two par ts n ext to it are called adjace n t

p ar ts, an d the remain in g two par ts are called Opp osit e p arts . For

a, b, c

,

7—B an d th ese ar e supposed to be ran ge d

e xample, if g—B is selected as t h e middle part, th en the adjacen t712

Then Napier ’s Rul es ar e t h e followin gsin e of the mi ddle part product of tan g en ts of adjacen t parts,sin e of t h e middle part product of cosin es of opposite parts.

0,an d th e Opposite parts are b an d 1

:-A.parts ar e a an d

2

67 . Napie r’

s Rules may be demon strated by shewin g that

th ey ag ree with the results already established. Th e followin g

table shews the required ag reemen t in the first column ar e g iven

the middle p ar ts, in the secon d column the results of Napier’

s

Rules , an d in the thir d column the same results expressed as in

Art . 62,with the n umber for referen ce used in th at Article .

SOLUTION OF RIGHT-ANGLE D TRIANGLES. 37

— c = tan —A tan cos c= cot A cot B

cos a cos b cos o = cos a cos b

B sin tan a tan cos B tan a cot c

cos b cos cosB cos 6 sin A

sin a t an b t an sin a tan b cot

sin a = cos Sin a z sin A sin c

sin b= tan tan a sin b= cot A tan a

sin b= cos sin b= sin B sin c

A sin tan bt an cosA z tan b cot c

cos a cos cosA cos a sin

Th e last four cases n eed n ot h ave been given , sin ce it is obvious

th at th ey ar e on ly repetition s of what h ad previously been g iven ;t h e seven th an d eig h th ar e repetition s of the fifth an d sixth

,an d

t h e n i n th an d ten th ar e repetition s of the th i rd an d fourth .

68. It h as been sometimes stated th at t h e meth od of t h e

precedin g Article is the on ly on e by which Napie r’s Rules can be

demon strated this statemen t,h owever

,is in accurate, sin ce besides

this meth od Napier himself in dicated an other method of proof inh is Mirifici Log arithmorum 0an 0n is De scrip tio, pp. 32 , 35 . Th is

we wil l n ow briefly explain ,


Le t ABO be a spherical tri an g le at 0 ; B

as pole describe a g r e at circle DE E G’

,an d with A as pole describe

ag reat circle HE E L,an d produce the sides of the orig in al trian g le

ABO to meet these g reat circles . Then sin ce B is a pole of DE E Gthe an g les at D an d 0 me rig ht an g les, an d sin ce A is a pole ofHE E L the an g les at H an d L ar e rig ht an g les . Hen ce the five

tr ian g les BAO, AE D,E E H, E KG,

E BB ar e all r ig h t-an g led an d

moreover it will be foun d on exam in ation that,althoug h t h e ele

men ts of these trian g les ar e differen t,yet th e ir circular par ts are

th e same . We will con sider,for example

,t h e trian g le AE D the

an g le E AD is equal to t h e an gle BAO t h e side AD is the complemen t of AB ; as the an g les at 0 an d G ar e rig ht an g les E is a

pole of 00 (Art . therefore E A is the complemen t of A0 ; as

B is a pole of DE the an g le BE D is a rig ht an g le, therefore the

an g le AE D is the complemen t of the an g le BE O,that is

,the

an g le AE D is the complemen t of the side B0 (Ar t . an d s

larly~t h e Side DE is equal to the an g le DBE ,

an d is therefore the

complemen t of the an g le ABC. Hen ce,if we den ote t h e elemen t s

of t h e trian g le ABO as usual by a,b,c,A

,B

,we have in the

t rian gle AE D the hypoten use equal to g—b, the an g les equal to

A an d g— a,an d t h e sides respectively opposite these an gles equal

Ban d g - c. Th e ci rcular p ar ts of ABD ar e th erefore the

SOLUTION OF RIGHT-ANGLED TRIANGLES . 39‘

same as those of ABO. Similarly the remain in g three of t h e five

rig ht-an gled tria n g les may be shewn to have t h e same circular

parts as the trian g le ABO h as.

Now take two of the theorems in Ar t . 65 , for example 1) an d

(3) th en the truth of the t e n cases comprised in Napie r’

s Rules

will be foun d t o follow from applyin g the two theorems in succes

sion to the five trian g les formed in the precedin g figure. Thus

this method of con siderin g Napier ’s Rul es r eg ards each Rule, n ot

as the statemen t of dissim ilar properties of on e t r ian g le, but as the

statemen t of similar properties of five allied trian g les .

69. I n Napier ’s work a fig ure is g iven of which that in the

precedin g Article is a copy, except that di fferen t let t ers ar e used

Napier briefly in timates that the truth of the Rules can be easily

seen by mean s of this fig ure, as well as by t h e method of in duct ionfrom con sideration of all the cases which can occur . Th e lat e

T. S . Davies,in h is edi tion of Dr Hutton ’

s Course of Ma th ema t ics,

drew atten tion t o Napier’

s own views an d expan ded the demon

st rat ion by a systematic examin ation of t h e fig ure of the precedin gArt icle.

It is however easy to evade t h e n ecessity of examin in g the

whole fig ure ; all that is wan t ed is to observe the con n ex ionbetween the tr ian g le AE D an d t h e trian g le BA0. For let a,

a

a3 ,

a4 ,

a5r epresen t t h e elemen ts of the trian g le BAO taken in

order,beg in n in g with t h e ~h ypot e n use an d om it tin g the r ig ht

an g le ; then the elemen ts of the trian g le AE D taken in order,

beg in n in g with t h e h ypoten use an d omi ttin g the rig h t an g le, ar e7 ? 7 T 7 T 7T

2a3, 5 2 2

a an d “2° If

,th erefore

,to ch aracterise

the former we in troduce a n e w se t of quan t ities Pu p 2 , p a, p 4 , p é ,

such that g , an d that p 3= a

3an d p 4

= a4 ,

then t h e origin al trian g le bein g ch aracterised by PI , p g , 17 3 , p 4 , P5 :

the se con d tr ian g le will be similar ly characterised by p a, p 4 , 295 ,

Pu p g. As the secon d tr ian g le can g ive r ise to a t hird in like

man n er , an d so on , we se e that every right-an g led trian gle is on e


of a system of five such tr ian g les which ar e all ch aracterised bythe quan t ities p l , p z, p 8 , p 4, p é ,

always taken in order,each

quan t ity in it s turn stan din g first .

Th e late R . L. E llis poin ted out th is con n exion between thefive t rian g les, an d thus g ave the true sig n ifican ce of Napier

’

s

Rules . Th e memoir con tain in g Mr E lli s ’s in vestig ation s, whi ch

was un published when the fir st edition of the presen t work ap

pe ar e d, wil l be foun d in pag es of Th e Ma th ematical an d

o th e r wr i tin g s of Robe rt Le slie Cambridg e, 1 863 .

Napier ’s own method of con side rmg h is Rules was n eg lectedby wr it ers on the subj ect un til the late T. S . Davies drew attention to it . Hen ce

,as we h ave already remarked in Ar t . 68

,an

erron eous statemen t was made r espectin g the Rules. For in

s tan ce,Woodhouse says, in h is Tr ig on ome try :

“ There is n o separate an d in depen den t proof of these rules A iry says

,in t h e

treatise on Trig on ometry in the E n cyclop azdia Me tr op olitan a

These rules are proved t o be true on ly by showin g that they comprehen d all the equation s which we have just foun d.

7 0. Opin ion s h ave difie r e d with respect t o t h e u tili ty of

Napier ’s Rules in practice. Th us Woodhouse says,

“ I n the whole

compas s of math emat ical scien ce th ere can n ot be foun d,perhaps,

rul es which more completely attain that whi ch is the properObj ect of rules, n amely

,facility an d brevity of computation .

”

(Tr ig on ome try, chap . X .) On the other han d may be se t the fol

lowin g sen ten ce from A iry’s Tr ig on ometry (E n cyclop ce dia Me tro

p oli ta n a):“ I n the opin i on of Delambre (an d n o on e was better

qualified by experien ce t o give an opin ion )these theorems ar e best

r ecollected by the practical calculator in th eir un con n ected form .

”

S e e Delambre’s Astron omi e , vol. I . p. 205 . Professor De Morg an

stron g ly obj ects to Napier ’s Rules, an d says (Sp h e r ical Tr ig on o

m e try, Art .

“There ar e certain mn emon i cal formulae called

Nap i e r’

s Ru le s of 0ircu lar Par ts, which ar e g en erally explain ed.

We do n ot g ive th em ,because w e ar e con vin ced that they on ly

create con fusion in stead of assistin g th e memory.

SOLUTION OF RIGHT-ANGLED TRIANGLES. 41

7 1 . We shall n ow proceed to apply the formulae of Art . 62

to the solution of rig ht-an g led tr ian g les . We Shall assum e thatthe g iven quan tities ar e subj ect to the limitation s which ar e stated

in Arts. 2 2 an d 2 3,that is

,a g iven side must be less than the

semi circumferen ce of a g reat circle, an d a g iven an g le less th an

two r ig ht an g les . Th ere will be six cases to con sider .

Her e we have from (5 )an d (2) of Art . 62,

t an b z tan c cos A,

cot B = cos c t an A,- sin a = sin c sin A.

Thus b an d B ar e determin ed immediately wi thout ambig uityan d as a must be of t h e same affection as A (Ar t . a also isdetermi n ed without ambigui ty.

It is obvious from t h e formulae of solution,th at

‘

in th is casethe trian g le is always possible .

If 0 an d A ar e both rig h t an gles, a is a righ t an g le, an d b an dB are in determin ate.

7 3 . Havin g g ive n a side b an d th e adjace n t cmg le A.

Here we h ave from (4)an d (6) of Art . 62,

_

t an b

cos z i’

Here 0,a,B ar e determin ed with out ambig ui ty, an d t h e t ri

an gle is always possible .

7 4. Havi n g g ive n th e two side s a an d b.

Here we h ave from ( 1) an d (4)of Art . 62,

cos c = cos a cos b,

cot A = cot a sin b, cot B = cot bsin a.

tan a z tan A sin b, cosB z cos bsin A.

Here c, A,B ar e dete rmin ed with out ambig uity, an d t h e t ri

an g le is always possible.

7 5 . Havin g g ive n th e h yp ote n use c an d a sid e a.

Her e we h ave from (3)an d (2) of Art . 62 ,

cos 0 t an a

cos Bcos a t an 0

42 SOLUTION OF RIGHT-A NGLED TRIANGLES.

’

Here b,B

,A ar e determin ed without ambig uity, sin ce A must

be of the same affection as a . It will be seen from th ese formulae

that th er e ar e limitation s of t h e data in order to in sur e a possibletrian g le in fact

,c must lie between a an d r: a in order that the

values foun d for cos b,cosB

,an d sin A may be n umer ically n ot

g reater than un i ty.

If 0 an d a‘

ar e rig h t an g les, A is a rig ht an gle , an d b an d B are

in determin ate.

7 6 . Havin g g ive n th e two an g le s A an d B.

Here we h ave from (5) an d (6) of Ar t . 62 ,

cos A cosB

s in B sin A

Here c,a,b ar e determin ed without ambig uity. There ar e

limitat ion s of the data in order to in sure a possible tri an g le. Fir st

7 ? 7 T—

2—A an d —

2- +A

,

cos c cot A cot B,

cos a

suppose A less than if,then B must li e between

n ext suppose A g reater than then B must lie between

7 T 7T

(7r —A)an d -

2 (7 r —A), th at I s, between A

7 7 . Havin g g ive n a sid e a an d th e Opp osi te armle A .

Here we have from (4)an d (6) of Ar t . 62 ,

sin a

sin A’

sin b= t an a cot A, sin B

Her e there is an ambig uity, as t h e parts ar e determin ed fromtheir Sin es . If s in a be less than sin A

,there ar e two values

admissible for c ; correspon din g t o each of these there will bei n g e n e ra l on ly on e adm issible value of b

,sin ce we must h ave

cos c cos a cos b,an d on ly on e admi ssible value of B

,Sin ce we

must have cos c cot A cot B. Th us if on e trian g le existswith the

g iven parts, th ere will be in g e n e ral two, an d on ly two, trian g les

with the g iven parts . We say in g e n e ral in the precedin g se n

t e n ce s,be cause if a =A there will be on ly

f'

on e trian g le,un less a

44 EXAMPLES.

4. Sin a tan 4A— sin b tan 4B ==sin (a — b).

5 . Sin (c a): sin bcos a tan 4B,

Sin (c'

6. If ABO be a spherical trian g le, rig ht-an g led at 0,an d

cosA = cos2a, sh ew th at if A be n ot a rig ht an g le b+ c = 47r or

gr , accordin g as b an d c are both less or both greater than 37 . If a , B be t h e arcs dr awn from t h e rig h t an g le respectively

perpen dicular to an d bisectin g the hypoten use c, shew th at

8. I n a trian gle,if 0 be a rig h t an gle an d D t h e mi ddle poin t

ofAB, sh ew th at

4 cosf -gSin”CD sin

”a sin

’ b.

9 . I n a. rig h t-an g le d trian g le, if 8 be the len gt h of t h e are

drawn from 0 pe rpen dicular to the h ypoten use AB, shew th at

cot 3 J(cot9a cot

9b).

1 0. 0AAlis a Spherical trian g le rig ht—an g led at AI

an d acute

an g led at A the ar c AlA2of a g reat circle is dr awn perpen dicular

t o 0A,then A

,A3is drawn perpen dicular to OA,,

an d so on sh ew

that ARA

”+ 1van ishes when n becomes in fin i te an d fin d the value

of cosAAlcos A

‘A2cos A

21 1

1 1 . ABO is a rig h t-an g le d spherical tri an g le, A n ot bein g t h e

righ t an gle sh ew that if A a,then c an d b ar e quadran t s.

1 2 . If 8 be the len g th of the are drawn from 0 perpen dicular

t o AB in a n y trian g le, shew that

cos 3 cosec c (cos’a cos

” b 2 cos a cos 6 cos c)5.

1 3. ABC is a g reat circle of a sphere AA’

,BB

'

,are arcs

of great circles drawn at righ t an gles t o ABC an d reckon ed pos1

EXAMPLES. 45

tive wh en th ey lie on t h e same side of it sh ew th at t h e con ditionof A

’

,B’

, 0’ lyin g in a g reat circle is

tan AA’

sin B0 t an BB’

sin 0A tan 00’

sin AB 0.

1 4. Perpen diculars are drawn from the an g les A,B

, 0 of an y

t ri an g le meetin g the Opposite sides at D,E,E respectively : sh ew

th att an BD t an 0E tan AE = tan DC t an E A tan E B.

1 5 . 0y ar e two g reat cir cles of a sph ere at rig h t an g les toe ach other

,P is an y poin t in AB an other g reat circle. 00 p is

the arc perpen dicular to AB from 0,makin g the an g le 0056 : a

W ith 090. PM,PN ar e arcs perpen dicular to 0m, 0y respectively

sh ew th at if 0M : x an d ON y,

cos a t an x + sin a tan y = tan p .

1 6. Th e position of a poin t on a sph ere, wi th referen ce to twogreat circles at rig ht an g les to each other as axes

,is determi n ed

by the portion s 0, (t of these circles cut off by great circles th rough

t h e poin t,an d th rough two poin ts on t h e axes

,each 3from th eir

poin t of in tersection : shew th at if t h e three poin ts (0,qS), (If),ct”

) lie on the same great cir cle

t an gb(tan 0’

- t an + t an q5’ (tan 0 — tan 0)+ tau gt (tan O— t an O

’

)= 0.

1 7 If a poin t on a sph ere be referred to two great circles atrig ht an g les to each oth er as axes

,by mean s of the portion s of

these axes cut off by g reat circles drawn throug h the poin t an dtwo poin ts on t h e axes each 900 from their in tersection

,sh ew th at

the equation to a great circle is

t an d cot a + tan ¢ cotB= L

1 8. I n a sph erical trian gle,if A an d 0

th at a +b .+ c7:2

.

( 46 )

VI. SOLUTION OF OBLIQ UE -ANGLE D TRIANGLE S.

7 8; Th e solution of oblique-ang led trian g les may be made in'

some cases to depen d immediately on the solution of r ig ht-an g ledtrian g les we will in dicate th ese cases before con siderin g the sub

je ct g en erally.

( 1) Suppose a trian gle to h ave on e of it s g iven Sides equal to’aquadran t . I n this case the polar trian g le h as its correspon din g"

an g le a rig ht a n g le the polar t rian g le can therefore be solved bythe rules of the precedi n g Chapter , an d thus the elemen ts of t h e

primitive trian gle become kn own .

(2) Suppose amon g the g iven elemen ts Of a trian g le th ere ar e

two equal sid e s or two equal an g le s. By drawin g an arc from t h e

vertex to the m iddl e poin t of the base, t h e trian g le is di vided in to

two equal rig h t-a n g le d trian g les ; by the solution of on e of these

rig h t an g le d tr ian g les the required elemen t s can be foun d.

(3) Suppose amon g the g iven elemen ts of a tr ian g le there

ar e two Sides, on e of which is t h e supplemen t of the other,or two

an g les, on e (

of wh i ch is the supplemen t of th e ot h er . Suppose, for

example,that b c = 1 r

,or else that B 0 1 r ; produce BA an d

B0 t o meet at B’

(see the fir st fig ure to Art . then t h e t rian g le

B’

AC h as two equal sides g iven ,or else two equal an g les g iven ;

an d by t h e precedin g case the solution of it can be made to depen d

on the solution of a r ig ht-an g led trian g le.

7 9 . We n ow proceed to the solution of oblique-an g led t ri

an g lesin g en eral . There wil l be six cases t o con sider .

80. Havin g g ive n th e th r e e side s .

cos a — cos b cos cHere we h ave cos A an d s imi lar formulae

S i n b Si n 0

for cos B an d cos 0. Or if we wish to use formulae suited t o log a

SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 47

rit hms, we may take the formula for the sin e, cosin e, or tan g en t of

half an an g le g iven in Art . 45 . I n selectin g a formula, atten tion

should be paid t o t h e r emarks i n Plan e Trig on ome try, Chap. X I I .

towards the e n d.

8 1 . Havin g g ive n th e th r e e an g le s.

cosA cos B cos 0Here we h ave cos a an d similar formulae

8 1 11 B 8 1 1 1 0

for cos b an d cos c. Or if we wish to use formulae suited to log a

r it hms,we may take the formula for t h e sin e

,cosin e

,or tan gen t of

h alf a side given in Art . 49.

There is n o ambiguity in the two precedin g cases ; t h e trian g le s

h owever may be impossible with the given elemen ts.

82 . Havin g g ive n two sid e s an d th e in clude d an g le (a, C, b).

By Napier ’s an alog ies

cos 4(a — b)tan 4(A + B)

t an 4(A—B) 442123001340 ;th ese determin e 4(A B)an d 4(A B), an d then ce A an d B.

sin a Sin 0

sin A

this case,sin ce c is foun d from it s Sin e

,it may be un certain wh ich

of two values is t o be g iven to it ; the poin t may be somet imessettled by observin g that the g reater side of a trian g le is oppos ite

to t h e'

g r e at e r an g le. Or we may determ in e 0 from equation ( 1)ofAr t . 5 4, which is free from ambig uity .

Th en c may be foun d from t h e formula sin c

Or we may determi n e 0, without previously determ in in g A an d

B,from the formula cos c cos a cos b sin a Sin bcos 0 this is

free from ambiguity. T hi s formula may be adapted to logarith mst hus

cos c = cos b (cos a ll- sin a tan bcos

48 SOLUTION OF OBLIQUE-ANGLED TRIANGLES.

assume t an 9 t an bcos 0 ; then

cos bcos (a 0)cos 0

th is is adapted to logarithms.

Or we may treat th is case con ven i en tly by resolvin g t h e t ri

g le in to the sum or differen ce of two rig h t-an g led trian g les.

From A draw the arc AD perpen dicular to 0B or 0B produced ;then

,by Art . 62 , t an 0D = tan bcos 0

,an d this determin es CD

,

an d then DB is kn own . Agai n,by Ar t . 62 ,

cos bcos c cosAD cosDB cosD

BCOS CD

this fin ds 0. It is obvious th at CD is wh at was den oted by 0 in

t h e former part of the Ar ticle.

By Ar t . 6 2 ,

tan At an 0 sin CD,an d tan AD= tan ABD sin DB ;

thus t an ABD sin DB t an 0 sin 0,

wh ere DB - : a — él or O- a,accordin g as D is on 0B or 0B pro

duce d,an d ABD is either B or the supplemen t of B ; th is for

mula en ables us to fin d B in depen den tly of A.

Thus, in the presen t case, th ere is n o real ambig uity, an d th etrian gle is always possible.

SOLUTI ON or OBLIQUE-ANGLED TRIANGLES. 44)

83 . Havin g g ive n two an g les an d th e in cluded side (A,c,B).

By Napie r’

s an alogies,

cos 5» (A B)

sin 5 (A B)tan % (a(A+3 )

tan % c ,

th ese determin e % (d b)an d % (d b), an d th en ce av

an d 6.

Th en 0 may be foun d from t h e formula sin 0sm

sfns

fl 0

th is case,sin ce 0 is foun d from it s sin e, i t may be un certain wh ich

o f two values is to be g iven to it ; t h e poin t may be sometimes

settled by observin g that the greate r an g le of a trian g le is Opposite

to t h e g reater side. Or we may determ in e 0 from equation (3) of

Art . 5 4,wh ich is free from ambig uity.

Or we may determin e 0 with out pr ew ously determi n in g a an d

b from the formula cos 0 cos A cos B sin A sin B cos c. Thisformula may be adapted to logarithms

,th us

cos 0 = cos B (— cosA + sin A tan B cos c);assume cot ct z tan B cos c th en

cos 0= cos B (— cosA+ cotqbsin A)cos B

zgi

p—qb)

;

th i s is adapted to logarithms.

Or.we may treat th is case con ven i en tly by resolvin g t h e t ri

an g le in to the sum or differen ce of two right-an g led trian g les .

From A draw t h e arc AD perpen dicular to 0B (se e the r ig ht

h an d figure of Ar t . th e n,by Ar t . 62

,cos c= cot ot DAB,

an d th is determ i n es DAB,an d th en CAD is kn own . Again

,

by Ar t . 62 ,

cosAD sin 0AD cos 0, an d cosAD sin BAD cos B ;cos 0 cos B

th ereforesin 0A1) sin BAD ;

th i s fin ds 0.

It is obvious that DAB is wh at was den ote d by 98 in t h e formerpart of the Article.

T. S . T.

5 0 SOLUTION OF OBLIQUE-ANGLED,TRIANGLES.

By Art . 62,

tan AD tan AO cos CAD,an d tan AD tan AB cos BAD ;

thus tan 6 cos 0AD t an 0 cos (I),

where CAD : A <1; th is formula en ables us t o fin d 6 in de pe n

de n t ly of a.

Similar ly we may proceed wh en th e perpen di cular AD falls on

0B p roduce d (se e the left-han d figur e of Art .

Th us,in the presen t case, there is n o real ambig uity ; more

over the trian g le is always possible.

84. Havzn g g zue n two sid es an d th e an g le opposi te p n e of th em

(a, b, A).

Th e an g le B may be foun d from the formula

6sin A

sm a

an d th en 0 an d 0 may be foun d from Napier’s an alog ies,

cos 1} (a b)tan i -B),

tan l ccos § (A

I n thi s case, sin ce B is foun d from it s sin e, th ere will sometimes

be two solution s ; an d sometimes the re will be n o solution at all,

n amely,when the value foun d for sin B is g reate r than un i ty. We

will presen tly return to this poin t. (Se e Art .

We may also determin e 0 an d 0 in depen den tly of B by for

mulse adapted to logar ithms. For,by Art . 44,

cot a sin b z cos b cos 0 + sin 0 00t A = cos b (003 0 +

cot Aassume tan g!)

cos b

cos bcos (0 — 96)cot a sin b z cos b 0)

therefore cos (0 005 (I) 00 17 a. t an 6 3

5 2 SOLUTION or OBLIQUE-ANGLED TRIANGLES.

A lso, by Art . 62 , tan AD = tan AO cosA th is fin dsAD. Then

cosAO cos 0D cosAD,

cos 0B cos 0D cosBD,

or cos 0B’

cos CD cosB’

D

cos A0 cos 0B cos 0B'

.

cos AD cos B!) cos B’

D

th is fin ds BD or B’

D.

th erefore

It is obvious th at AD is wh at was den oted by 0 in t h e former

part of the Article.

85 . Havin g g ive n two an g les an d th e side opposite on e of th em

(A,B,a).

Th i s case is an alog ous.

to th at immediately precedin g , an d

g ives r ise to t h e same ambig uities . Th e side 6 may be foun d from

t h e formula

sin 6sin B sin a

sin A

an d th en 0 an d 0 may be foun d from Napier ’s an alog ies,

— b)

+B)cos é (A — B)

cot (A B),

tan (a b).

We may also determin e 0 an d 0 in depen den tly of b by formulaeadapt ed to log ar ithm s. For

cos A : - cosB cos C + sin B sin 0 cos a

cos B cos 0+ tan B sin 0 cos a),

assume cot <1) tan B cos a th us

cosB sin (0 — ct)1

cos A= cosB (8)q

therefore si n (0 gt)

SOLUTION or OBLIQUE-ANGLED TRIANGLES. 5 3

from th is equation 0 oS is to be foun d an d th en 0. Sin ce 0 cf;is foun d from it s Si n e there may be an ambig uity. Again

,by

Ar t . 44,

co t A sin‘

B= cot a sin c - cos c cOSB= cosB cos c +

assume cot 0

cot A sin B cos B cos c sin c cot 9)COS

Bil

l

}

?6)5

th erefore sin (c 9) cot A tan B Sin 0

from this equation c 9 is to be foun d, an d th en 0. Sin ce 0 0 is

foun d from it s sm e there may be an ambig uity. AS before,it may

be shewn that these results ag ree with those obtain ed by resolvin gthe trian g le in to two rig ht-an g led trian gles ; for if in t h e trian g le

AOB’ the ar e CD be d rawn perpe n dicular to AB

’ th en B’

0D

will 96, an d B’

D 0.

86 . We n ow r etur n to t h e con sideration of t h e ambig uity

wh ich may occur in t h e cas e of Ar t . 84,wh en two sides are g iven

an d t h e an g le opposite o n e of th em . Th e discussion is somewh at

tedious from it s len gth,but presen ts n o difficulty.

Before con siderin g t h e problem gen erally,w e will take t h e

particular case in wh i ch a : b ; th en A must =B. Th e first an d

third of Napier ’s an alog ies give

cot %0 = tan A cos a,

tan § c = tan a cos A ;

n ow cot 150 an d t an l

zwmust both be p osi tive , so th atA an d a. must

be of the same affection . Hen ce, wh en n z b,th ere will be n o

Solution at all,un less A an d d ar e of t h e same affection

,an d th en

there will be on ly on e solution ; except when A an d a ar e both

rig ht an g les, an d then cot %0 an d t an fie are in determin ate,an d

there i s an in fin i te n umber of solution s .

We n ow proceed to t h e gen eral discussion .

If sin bsin A be greater than Sin a,there I s n o trian gle wh ich

satisfies the given con di tion s ; if sin bsin A is n ot g reater th an

5 4 SOLUTION or ~OBLI Q UE ~ANGLE D TRIANGLES.

0 O 0 Sin 8.

Asm a

,t h e equati on sm B

b mfurn ish es two value s of B

,S in a.

whi ch we wil l den ote by B an d B'

,so th at B

’

1r B ; we will suppose that B is t h e on e wh i ch is n ot g reater than the oth er .

Now,in order that th ese values of B may be admi ssible

,it is

n ecessary an d sufficien t that t h e value s of cot %0 an d of tan 5 0should both be positive, th at is, A—B an d a b must h ave thesame Sig n by t h e secon d an d fourth of Napier

’

s an al og ies. We

have th erefore to compare t h e sign ofA - Ban d the Sign of A B'

with that of a b.

We will suppose th at A is less than a rig h t an gle, an d separate

t h e correspon din g discussion in to th ree cases.

I. Le t b’

be‘

le ss th an 352

( 1) Le t a be less th an b,th e formula Sin B Si_n_ h .

sin Amakessin a

B g reater th an A, an d d. for tiori B’

g reater than A. Hen ce th ere

ar e two solution s .

(2) Le t a be equal to“

b th e n ”

th ere‘

I s on e solution,as pre

viously sh ewn .

(3) Le t a be greate r th an b ; we may have then a b less thanar or e qual to 7 : or greater th an 7 r. If a b is less than 7 r

,then

si n a is g reater t h an Sin b ; thus B is less than A an d therefore

admissible,an dB

'

is greater th an A an d in admissible. Hen ce there

is on e solution . If a + b is equal to 7 r, then B is equal to A,an d

B'

g reater than A,an d both are in admissible. Hen ce ther e is n o

solution . If a + b is g reater than 7 r, then sin a is less than sin b,

an d B an d B’

ar e both greater t h an A,an d both in admi ssible. Hen ce

th ere is n o solution .

II. Le t ~b be equal to“

1;

( 1) Le t a be less th an b; th en B an d B’

are both g reater th an

A,an d both admissible. Hen ce th ere ar e two solution s .

(2) Le t a. be e qual to b ; th en th ere is n o solution , as pr e

viously sh ewn .


( 3) Let'

a‘

h e -greater th an b‘

th en sin‘

a is less th an sin b,an d

B an d B’

ar e both greater than’

A,an d in admissible. Hen ce there

is n o solution .

III. Le t b be greater th an

( 1) Le t a be less th an 6 we may h ave th en a + b less than

7 : Or equal to 7 : Or g reater than 7r. If a + b is less than r,then

sin a is less than Sin b,an d B an d B

’

ar e bOt h g reate r than A an d

both admissible. Hen ce there ar e two solution s . I f’

a +b is equalto qr

,then B is equal to A an d in admissible

,an d B

’

is g reater

th an A an d admissible. Hen ce ther e is o n e solution . If a b

is g reate r than 7 r,th en sin a is g reater than Sin b ; B is less

than A an d in admissible,an d B

’

is g reater than A an d admissible .Hen ce there is on e solution .

( 2) Le t a be equal to b ; th en th ere is n o solution,as pre

viously shewn .

(3) Le t a be greater th an b ; th en sin a is less th an sin b,

an d B an d B’

ar e both g reater than A an d bot h in admissible .

Hen ce th ere is n o solution .

We h ave th en t h e followin g results wh en A is less th an a

r ig h t an g le .

a b two solution s,a b on e solution

,

a b an d a b 7r on e solut ion,

a > b an d a + b= 7r or > 7r n o solution .

1 : a b two solution s,2 ( a b or a b n o solution .

a b an d a 3& b 7 1' two solution s

,

a b an d a b W or ir on e solution ,

a b or b n o solution .

5 6 SOLUTION or . OBLIQUE-ANGLED TRIANGLES .

‘

It must be r emembe r e d,~h owe ve r

,th at in the cases in wh ich

two solution s are in dicat ed, th ere will be n o solution at all if

sin a be less th an sin bsi n A.

I n t h e same man n er t h e cases in wh i ch A is equal to a rig h tan g le or g reater than a right an gle may h e di scussed, an d thefollowin g results obtain ed.

Wh e n A is equal to a rig h t an g le ,

a b an d a + b= 7 r or > 7 r

 b

b in fin ite n umber of solution s.

a b an d a b 7 : on e solution,

a b an d a b 7x or 7 : n o solution,

a b or a b n o solution .

Wh e n A is g r e ate r than a r ig h t an g le ,

a b an d a + b= 1 r or < 7 r on e solution,

a b an d a b 7 : two solution s .

b or a b n o solution,

w w

a < b an d a + b= 7r 0r < 7r

a b

a b two solution s .

As before in the cases in wh i ch two solution s are in dicated,

there will be n o solution at all if sin a be less th an sin 6 sin A.

It will be seen from th e above in vestig ation s th at if a li esbetween b an d i r —

'

b, there will be on e solution ; if a does n ot li e

between b an d 7 r— b eith er th ere are t wo solution s or there isn o solution ; th i s en un ciation is n ot mean t to in clude t h e cases in

wh ich a = b or = 7r — b.


’ 87 . Th e results of th e precedin g Article may be illustrated by

Le t ADA’

B be a great circle ; suppose PA an d PA’

t h e

projection s on the plan e of th is circle of arcs wh i ch ar e eachequal to b an d in clin ed at an an g le A to ADA’

; le t

PE be th e projection s of t h e least an d g reatest distan ces of P

from t h e gre at circle (se e Art . Th us the fig ure supposes

A an d 6 each less th an

If a be less th an th e are wh ich is represen ted by PD th ere isn o trian g le if a be between PD an d PA in mag n i tude, th ere ar e

t wo trian g les, sin ce B will fall o n ADA’

,an d we h ave two trian gles

BPA an d BPA’

; if a be between PA an d PH there will be on ly

on e trian g le, as B will fall on A’H or AH’

,an d the trian g le will be

either APB with B between A’

an d H,or else A’

PB with B be

tween A an d H’

; but th ese two trian g les are symmetrically equal(Art . if a be greater than PH there wi ll be n o trian gle.

Th e fig ure will easily serve for all th e cases th us if A is greater

th an 35,we can suppose PAE an d PA

’

E to be equal to A ; if

b is greater th an we can take PH an d PB7 to represen t b.

5 8 SOLUTION or OBLIQUE -ANGLE D TRI ANGLES.

88. Th e ambig uities wh ich occur i n t h e last caSe in‘

t h e solution of oblique-an g led trian g les (Art . 85 )may be di scussed in t h esame man n er as those in Art . 86 or

,

’

by mean s of the polartrian g le, the last case may be deduced from th at of Ar t . 86.

EXAMPLES.

l . Th e Sides of a trian g le are an d 7 5°respectively

fin d t h e Sin es of all the an gles.

sin (s — c)Sin s

wh en a side,an adjacen t an g le, an d t h e sum of t h e oth er two

sides are g iven .

2 . Sh ew th at tan §A tan §B Solve a trian g le

3 . Solve a trian g le h avin g g iven a Side,an adjacen t an gle,

an d t h e sum of the oth er two an g les.

4. A trian g le h as the sum of two Sides equal to a semi a r

cumfe re n ce : fin d t h e are join in g t h e vertex with t h e mi ddle of

t h e base.

5 . If a, b , c ar e kn own ,6 bein g aquadran t, determin e the

an gles : shew also that if 8 be the perpen di cular ‘

on c from th e

Opposite an gle, cos”8 cos9a cos

gb.

6 . If on e side of a Spherical trian g le be divided in to fourequal parts, an d 9 ,

9 , 9 9 be the an g les Subten ded at the Oppo8 ’ 4

site an g le by the par tg

s taken I n order, Sh ew that

(91 sin 9,sin 9

,Sin (93 9

4)sin 91 sin 93 .

7 . I n a sph erical trian gle if A =B 20, sh ew that

90 6

8

2 2

00 CIRCUMSCRIBED AND INSCRIBED CIRCLES:

Th e value of tan 'r may be expressed in various forms th usfrom Art . 45

,we obtain

A

2 sin 8 Sin (3 a)

substitute th is value in th us

sin (8 a)sin (8 0»sin s J

. (Art .

sin (s —% a}

— cos %

sin i a cos §asin éA

sin a sin §B sin § 0 0

sin gA

sin § B sin § 0cos §A

{cos i (B 0) cos 5 (B (Art 5 4)

therefore from ( 1) tan r

h en ce, by Art . 5 1,

J{ cos S oos (S —A)cos (S —B)cos (32 cos éA cos § B cos § 0

N

2 COS %A COS %B COS % C

It may be sh ewn by common trigon ometrical formulae th at

4 cos 5Acos £B cos §0= cos S + cos (S —A) cos (S

h en ce we h ave from (4)

S + oos (S—0)

CIRCUMSCRIBED AND INSCRIBED C I RCLE S . 61

90. Tofin d th e an g ular radius of th e small circle de scribed

so as to touch on e side of a g ive n tr ian g le , an d th e oth er side s

Le t ABO be t h e trian g le ; an d suppose we require t h e rad ius

Of t h e small circle which touch es B0,an d AB an d A0 produced .

Produce AB an d A0 to meet at A’

; then we requir e t h e radius of

t h e small circle in scribed in A ’B0

,an d the sides of A

’

B0 ar e a,

7 r 6,7t 0 respe ctively. Hen ce if r

lbe the required radius

,an d

8 den ote as usual t} (a b c), we h ave from Art . 89,

At an r

l= t an

2~ sm s

From th is result we may derive oth er equivalen t forms as in

t h e precedin g Article ; or we may make us e of those forms im

mediate ly,observin g that the an g les of t h e trian g le A

’

BC ar e A ,

w —B, 7 r — 0 respectively. Hen ce 8 bein g an d S

bei ng 22 (A B 0)we sh all obtain

Sin s sin (s — b)sin (s — c)tan r,

8 1 11 (s a)

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

J{ cos S cos (S —A)cos (S —B)cos (S2 00s §A sin § B sin § 0

.

N

Acos fiA Sl n fiB sn l fic

- cos S - cos (S —A)+ cOS (S — 0)

62 CIRCUMSCRIBED AND INSCRIBED CIRCLES.

Th ese results may also be foun d i n depe n den tly by bisectin gtwo of the an g les of the trian g le A

'

BC, so as to , determin e t h epole of the small circle

,an d proceedin g as in Art . 89.

9 1 . A Circle wh ich touch es on e side of a trian g le an d t h e

other Sides produced is called an e scr ibe d circle ; th us there ar e

three escribed circles belon g in g to a g iven trian g le. We may

den ote the radii of the escribed circles which touch 0A an d AB

respectively by r,an d r

, ,an d values of tan r

san d t an r

3may

be foun d from what h as been already g iven with respe ct tot an r

lby appropriate ch an g es in t h e lett ers wh ich den ote t h e

Sides an d an g les.

I n the precedin g Article a t rian g le A’

B0 was formed by prodacin g AB an d A0 to meet ag ain at A

’

; similarly an other trian g lemay be formed by p roducin g B0 an d E A to meet ag ai n , an d

an oth er by producin g 0A, an d 0B t o meet ag ain . Th e origin altrian g le ABO an d the thre e formed from it have been called

associated trian g le s , ABO bein g the fun damen tal trian g le. Thusthe in scribed an d escribed circles of a g iven trian g le ar e t h e sameas the circles in scribe d in the system of associated trian gles ofwh ich t h e given trian gle is t h e fun damen tal t li an g le .

92 . To fin d th e an g ular radius of th e

about a g ive n trian g le .

Le t ABO be t h e given trian g le ; bisect the sides 0B, CA at

D an d E respect ively,an d draw from D an d E arcs at rig ht an g les

to 0B an d 0A respectively, an d le t P be t h e in tersection of these

CI RCUMS 63

arcs. Then P will be t h e pol

ABO. For draw PA,PB

,P0 ; th en from t h e rig ht-an g led

t rian g les POD an d PBD it follows that PB P0 ; an d fromthe rig ht-an g led trian g les POF an d PAE it follows that PA a: P0

hen ce PA PB P0. Al so the an gle PAB the an g le PBA,

t h e an g le PBO the an g le PCB,an d

”

the an g le POA the an g le

PAO ; therefore PCB A 5 (A B an d PCB S A.

Le t P0 R.

Now tan CD t an 0Pc os POD, (Art .

th us .tan iy a = tan B cos (S —A),

th erefore tan B51 3—1 15—6

Th e value. of tan B may be expressed in various forms - th us

if w e substitute for tan g from Art . 49,we obtai n

cos St an R

(S —A)cos (S — B)cos (S — 0)

Again cos (S —A)= cos (B + 0) —A}

Sin A cOS A

ios 0

}{cos % (b c) cosA» (b (Art .

sin Al

008 3acos é bcos 2 0 ,

th erefore from'

(1)sin a

tan RSin A 008 é

'

bcos 5 c0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Substitute in t h e last expression t h e value of Sin A fromAr t . 46 thus

2 sin § a sin § b sin % ctan BSi n s sin (s — a)sin (s — b)Sin (s — c)}

Z Sin t} a sin % bsin % c

64 ~CIRCUMSCRIBED AND INSCRIBED CIRCLES.

It may be shewn,by common trig on ometrical formulae, th at

4 Sin g a sin 5 b sin % c = sin (s — a)+ sin (s b)+ sin (s — c)— Sin s ;

h en ce we have from (4)

tan B (s — c)—sin s

93 . To fin d th e an g ula r radi i of th e small circle s described

rou n d th e tr ian g les associa te d wi th a g ive n fun dame n tal trian g le .

Le t RIden ote the radius of the circle described roun d t h e

trian g le formed by producin g AB an d A0 to mee t ag ain at A’

;

Simi larly le t R,an d B

3den ote t h e radii of t h e circles described

roun d the oth er two trian g les which ar e similarly formed. Thenwe may deduce expression s for t an R n

t an B2 ,

an d t an R3from

those foun d in Ar t . 92 for tan B. Th e sides of the trian g le A'

BC

ar e a ,i r — b

,7 r — c

,an d its an g les ar e A

,7r — B

,i r—0 ; hen ce if

s an d we shall obtain from

Ar t . 92

tan Bl

cos (S A)cos S cos (S B)cos (S 0)

sin § aSin A Sin é b Sin é w

sm asin (s — a)Sin (s — b)sin (s — c)}

tan R sin s — Sin (s — b)+ sin (s — c)

Similarly we may fin d expression s for tan B,an d tan B

3.

94. Man y example s may be proposed in volvin g properties of

the circles in scribed in an d described about the associated trian g le s .

We will g ive on e that will be of use h ereaft er .

CIRCUMSCRIBED AND INSCRIBED CIRCLES. 65

To prove th at

(cot r tan B)” £7? (Sin a sin b sin c)

“1 .

W e h ave

4n’z 1 — cos9 a - cos

°b

th erefore

(Sin a + sin b Sin c)’ 4n

2

2 l + sin a sin b+ sin b sin c + sin c sin a — cos a cos b cos

Also cot r -l- tan R5

1

3-

3

sin s + sin (s — c)

an d by squar in g both members of th is equation t h e required

re sul t will be Obtain ed. For it may be sh ewn by r e duction th at

Sin2s + sin

2

(s — a)+ sin2

(s — c)= 2 — 2 cos a cos bcos c,

sin s sin (s — a)+ sin s sin (s - b)+ sin s Sin (s — c)

+ Sin (s - a)sin (s — b)sin (s — c)sin (s — a)

sin a sin b+ sin bsin c + sin c sin a.

Similarly we may prove th at

(cot r , — t an B)2 — Sin a)

21 .

95 . I n t h e figure to Art . 89, suppose DP produced th rough

P t o a poin t A ’such th at DA’

is a quadran t,th en A’

is a pole of

B0, an d PA’

g— r similarly, suppose E P produced th rough P

t o a poin t B’

such th at E B ’

is a quadran t, an d FP produced

throug h ,P t o a poin t 0’

such th at F0’is a quadran t. Th en

A’

B’

O’is t h e polar trian gle of ABC, an d PA

’z PB

’: PO

’

3'

Th us P I s t h e pole of th e small circle d e scr ibed roun d t h e polar

trian g le, an d t h e an gular radius of t h e small circle described roun dt h e polar trian gle is t h e compleme n t of t h e an gular radius of the

66 CIRCUMSCRIBED AND i NSCRI BE D CI RCLES. »

small circle in scribed in the pr imitive trian g le. An d in like -man

n e r the poin t which,

is the pole of the small circle in scribed in

t h e polar tr ian g le is also the pole . of t h e small”

circle described

roun d the primi tive trian g le, an d the an g ular radii of the two

circles ar e complemen tary .

EXAMPLES.

I n th e followi n g example s th e n otation of th e Ch apter

Sh ew th at in an y trian g le t h e followin g relation s h old con

t ain e d in E xample s 1 to 7

1 . Tan rltan r

gtan r

az tan r sin

’s.

2 .

t an B3+ cot

3.

cot9r + cot

2rIcot”z r

,+ cot

2

co t rlcot r

,cot r

3cot r

L

5 . Cose cgr z cot (s—a)cot (s—b) cot (s- b)cot (s—c) cot (s- c)(s- a).

6 . Cosec2 r1

cot (s b)cot (s —c)— cots cot (s b) -cot s cot (s—c).

7 . Tan Bltan tan B

sz tan B se c

’S .

8. Shew that in an equilateral trian g le tan R 2 tan r .

9. If ABO be an equilateral sph e rical trian g le, P t h e pole of

th e cir cle circumscribin g it , Q an y poin t on the sphere, Sh ew th atcos QA cos QB cos Q C 3 cos PA cosPQ .

1 0. If three small circles be in scribed in a spherical trian g le

havin g each of its an g les so that each touches the other two

as well as two sides of the trian g le, Shew that the radius of eachof the small circles an d that the cen tres of the three small

circles coin cide with th e an gular poin ts of t h e polar trian gle.

68 AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS.

97 . Tofin d th e ar ea of a Sp h e r ical Trian g le .

Le t ABO be a sph eri cal trian gle produce th e arcs wh ich form

it s sides un til they m eet ag ain two an d two,wh i ch wil l h appen

wh en each h as become equal t o t h e semi-circumferen ce . Th e

trian g le ABO n ow forms a part of three lun es, n amely,ABDOA,

BOE AB,an d 0AFBO. Now the trian g les 0DB an d AFB are

subt en ded by vertically opposite solid an g les at 0, an d we wi ll

assume th at their areas ar e equal therefore the lun e 0AFBO is

equal t o the sum of t h e two tr ian g les ABC an d ODE . Hen ce if

A,B

, 0 den ote t h e circular measures of t h e an gles of t h e trian g le,we h ave

trian g le ABC BGDC lun e ABDOA 2Ar9

,

trian g le ABO AHE O lun e BCB'

AE 2Br’

,

trian gle ABC trian gle 0DE lun e 0AFBO 20r’

h en ce, by addition ,

twice trian gle ABO surface of h emisph ere 2 (A B 0)

th erefore t rian g le ABC (A B c t ry-2

Th e expression A B 0 7 : is called t h e sp h eri cal excess of

t h e trian gle ; an d Sin ce

AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 69

the result Obtain ed may be thus en un ciated : th e ar e a of a sp h e r ical

tr ian g le is th e'

same fraction of half th e surface of th e sp h e r e as th e

sp h e rica l excess is of four r ig h t an g les .

98. We h ave as sumed, as is usually don e,th at t h e are as of

t h e tr ian gles 0DE an d AFB in t h e precedin g Article are equal.

Th e trian g les ar e , h owever, n ot absolutely equal,but symme tr i

cally equal (Ar t . so th at on e can n ot be made to coin cidewith t h e oth er by superposition . It is, h owever, easy to de com

pose two such trian gles in to piece s wh ich adm it of supe rposition,

an d th us to prove that their ar eas ar e equal. For describe a

small circle roun d each , then t h e an gul ar radii of th ese circles

will be equal by Ar t . 92 . If t h e pole of t h e circumscribin g circle

falls in side each trian g le, th en each trian g le is t h e sum of three

isosceles trian g les , an d if the pole falls outside each trian g le, then

each trian g le is th e excess of two isosceles trian g les over a th ird

an d in each case t h e isosceles trian g les of on e se t ar e respectively

absolut e ly equal to t h e correspon din g isosceles trian gles of t h e

oth er se t .

99. Tofin d th e area of a sp h e rical p olyg on .

Le t n be t h e n umber of Sides of the polygon, 2 t h e sum of all

its an gles. Take an y poin t with in t h e polygon an d j oin it withall t h e an gular poin ts th us t h e figure is divided in to n tr ian gles .

Hen ce,by Art . 97 ,

area of polygon (sum of t h e an gles of th e trian gles n i t )r“

,

an d t h e sum of t h e an gles of t h e trian gles is equal to 2 togeth er

with t h e four righ t an gles wh ich ar e formed roun d t h e commonvertex ; th erefor e

area of polygon 2 (n 2) r".

Th is expr e ssion is true even wh en t h e polygon has some of it san gles g reater than two rig h t an gles

,provided it can be de com

posed in to trian gles, of wh ich each of t h e an gles is less th an tworigh t an gles.

70 AREA or A SPHERICAL TRIANGLE. SPHE RICAL EXCESS.

1 00. We sh all n ow g ive some e xpr e ssion s’

for’

ce i'tain t rig on or

metrical fun ction s of the sp h e r ical exce ss of a tri an g le. We den ote

t h e spherical excess by E ,so that E r: A + ~B 0 qr.

1 01 . Cag n o li’

s Th e or em . To shew that

{Sin s sin (s — a)sin (s — b)sin (s — c)}

- ll (a - C »

sin l (A+ 13)Sin 21 0— Cos i (A+B)COS %0

sin % EJ

os (a b) cosNa (Art . 5 4)

sin 0 sin é a sin §~ b008 5 0

2

sin a sin b' (s a)Sin (s — b)sin (8 c)}

Sl n s sin (s —a)Sin (s — b)Sin (s — c)}2 cos § a cos § bcos § c

102 . Lh uilie r’

s Th e or em. To Sh ew that

(s —a)tan a(s b)tan % (s

Sin i (A+B + 0 — 7 r)— 7r)

sin % (A+B)— Si n é» (7 r — 0)+B)+ cos % (7r —0)

— cos % 0

cos % (a — b)— cos % c COS %~C(Ar t .

Hen ce,by Art . 45 , w e obtain

Tan i E

(Plan e Trig . Art .

— c) sin (s — a)sin (a— b)

— a)tan % (s — b)tan g (s

ARE AO RA SPHE RI CAL TRIANGL E . SPHERICAL EXCESS . 7 1

1 03 . We may obtain man y oth er formulae in volvin g trig on ome t rical fun ction s of th e sphe rical e xcess. Thus

,for example

,

(A4B)

+B)cos 5§ 0

e a Sm9%0 + cos zl- (a se c%0, (Art .

cos aa cos lab (cosg

-é 0+ Sin ”%0)

— sin 2 % 0) se c é c

Again,it was shewn in Art . 1 01

,that

Sin

sin %a sin “li bSin 01 1?th erefore tan2

cos %a cos %b sin —% a sin é—bcos 0

Again,w e h ave from above

cos § E = cos § a cos'

se c % c

( 1 + cos a)( l + cos b)+ SI n a S I n b cos 0

4 cos §~ a cos § bcos 32m

1 + cos a + cos b+ cos c — l

l . _1_ LL _1_ _1_4 cos2a cos

zbcos % c 2 cos

2a cos

zbcos

zc

I n (3)put 1 2 Sinz

i E for COS % E ; th us

By ordin ary developmen t w e can Shew that the n umerator of

t h e above fraction is equal to

— b)sin % (s - C)5

7 2 AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS .

th e refore

cos § a cos § bcos § c

Similarly

co s 5a cos %b cos §

Hen ce by division w e obtain Lh uili e r’s Th eorem.

Sin 0 00t % E — COS C

” m osin l a sin l bai n o

cot é a cot 5b ;

th erefore,by Art . 1 01

2 Sin % a sin % b cos § c

Again , COS (0 - % E ) cos 0 cos $E + sin 0 Sin 5B

4 cos § a cos § bcos § c

( l + cos a)( 1 + cos b)cos 0 + sin a simb4 cos § a cos § bcos § c

sin’0sin 5a sin %bse c lz c

se c l c

4 Sin % a sin % b cos § c

cos c — cos a cos b+ ( l — cos a)(1 — cos b)4 sin § a sin § b cos § c

a

AREA OF A SPHERICAL TRIANGLE. SPHERICAL EXCESS. 7 3

From th is result w e can deduce two oth er results, in th e

same man n er as (4) an d (5) w e re deduced from (3) or we may

observe th at the righ t-h an d m ember of (6) can be obtain ed from

th e righ t-h an d member of (3) by writin g 7 r — a an d i r—b for

a an d b respectively, an d th us we may deduce th e results more

easily. We Shall h ave th encos % s sin % (s — c)

- t E )sin aa sin t b e os t e

2 1Sin % s cos % (s — c)COS ( 2 0 4

E )Sin % a sin % bcos % c

E XAMPLES.

1 . Fin d t h e an gles an d sides Of an equilateral trian g le wh ose

area is On e -fourth of th at of t h e sph ere on wh ich it is described.

2 . Fin d t h e surface of an equilateral an d equian gular Sph e

rical polyg on of n sides,an d determin e t h e value of each of t h e

an gles wh en t h e surface equals h alf t h e surface of th e sph ere.

E A Z3,

2,sh ew t h at E - cos

9.

4. If t h e an gle 0 of a sph erical trian gle be a righ t an gle,

Sh ew th at

5 . If th e an gle 0 be a righ t an gle,Sh ew th at

3 . If a = b

Si n’20 Sin 2 a 811 1

2b

cos Fcos 0 cos a cos b

2

6 . If a b an d 0 sh ew th at tan Esm a

2’

2 cos a

7 . Th e sum of t h e an gles in a righ t-an gled trian gle is lesst h an four righ t an gles.

8. Draw th rough a g iven poin t in t h e Side of a Sph ericaltrian gle an arc of a great circle cuttin g off a g lve n part of th etrian gle.

7 4 EXAMPLES.

9 . I n a Sph erical tr ian gle if COS 0 — tan“tan -i)

2)

0 z A+B.

1 0. If t h e an g les of a Sph erical triangle be togeth er equal tofour rig h t an g les

cos”

a cos2

ga b cos’3

i“ c 1 .

1 1 . I f r r, ,

r,be t h e radii of three small circles of a

Sph ere of radi us r which touch on e an other at P, Q, R,

an d

A, B, 0 be t h e an g les of t h e Sph erical trian g le formed by j oin in gtheir cen tres

,

area PQR (A cos rlB cos r

,0 cos r

37 7 )r

’.

1 2 . Sh ew that

% F sin (A — 5 E )

1 3. Given two Sides of a spherical trian g le, determi n e wh enthe ar ea is a

,maX imum .

1 4. Fin d the area of a reg ular polyg on of a given n umber of

Sides formed by arcs of g reat circles on the surface of a Sphere"

;

an d h en ce d educe that, if a be t h e an g ular radius of a small

circle, it s area is t o that of th e whole sur face of the sphere as

versin a is to 2 .

1 5 . A,B

, 0 ar e t h e an g ular points of a Sph erical trian g le ;A

'

,B’

, 0’ar e the middle poin ts of the respectively opposite sides.

If E be t h e spherical excess of the trian g le, shew that

cosA~

’B’

COS B’

0’

cos 0’

A’

cos g c cos g a cos § b

‘

1 6. If on e of the arcs of g reat circles wh i ch join t h e mi ddle

poin ts of the Sides of a spherical trian g le be a quadran t, shew

that the other two ar e also quadran ts.

7 6 ON CERTAIN APPROXIMATE FORMULAi .

— c),

cos g bcosfic= cos°

i (b+ c)— Sin’

i (b— c);

th erefore

— c)

l — sin ’ i (b

th erefore

9 sin A ( 1 - cosA)sin“

1 (b c) + cosA)sin’

i (b- c),

th erefore 9 = — cot %A sin’

i (h —c).

Th is gives th e circular me asur e of 9 ; th e n umbe r of se con ds int h e an gle is foun d by dividin g th e circular measure by t h e circularmeasure of on e se con d, or approximately by t h e sin s of on e secon d

(Pla n e Trig on ome try , Art . If the len gths of t h e arcs correspon din g to a an d b respectively be a. an d B, an d r th e radius of t h e

sph ere, we h ave“

an dB

as th e cir cular measures of a an d br

respectively ; an d th e len gth s of t h e sides of t h e ch ordal trian gle

2 r sin23; an d 2 r sin 2

6;

respectively. Th us wh en t h e sides of

t h e sph erical trian gle an d t h e radius of t h e sph ere ar e kn own,we

can calculate t h e an gles an d Sides of t h e ch ordal trian g le.

1 06. Legen dre’s Th eorem . If th e sid e sqf a sp h e r ical trian g le

be small compar ed with th e radius of th e sph e r e , th e n e ach an g le

qf th e sph e r ical trian g le exce eds by on e th ird of th e sp h e r ical ea:

ce ss th e corr espon d in g an g le of th e plan e tria n g le , th e side s ofwh ich are of th e same le n g th as th e arcs of th e sph e rical tr ian g le .

Le t A,B

, 0 be t h e an g les of th e sph erical trian gle ; a, b, c

t h e sides ; r th e radius of t h e sph ere ; a, B, y t h e len g th s of t h e

a B rarcs wh ich form th e side s

,so th at

r

are t h e circular

measur es of a,b,0 respect ively. Th en

ON CERTAIN APPROXIMATE FORMULAZ . 7 7

cos a—cos bcos c

sin bsin c

COS a = 1 c o o“OW2 73 2M

as

+r

Similar expre ssion s h old for cos b an d sin b, an d for cos can d sin 0 respectively. Hen ce

,if we n eglect powers of t h e 01r

cular measure above t h e fourth , w e h ave

1al

“

2 73 2 474

B4

_

74_ 6

2

72

)

14 4

6“ Bl-

7 B

le t A", B’

, C’

be th e an gles of th e wh osea, B, 7 re spe ctively 3 th en

B’+ 7

’

2

cosA= e osA’

cosA'

Suppose A A'

03 th en

cos A cosA'

9 sin A'

approximately 5

th erefore 9 M S

6r2

3r

7 8,

ON CERTAIN APPROXIMATE FORMUL E .

where S den otes t h e area Of~th e plan e trian g le wh ose sides ar e

a, B, 7 . Similarly

B =B’+ an d C = C

’

+

hen ce approximately

Sth erefore

7,1s approximately equal to t h e sph erical

sph erical trian gle,an d th us t h e th eorem is established.

It wil l be seen th at‘

in th e above approximation t h e area of

the spherical trian g le is con sidered e qual to t h e area of t h e plan e

trian gle wh ich can be form ed wi th sides of t h e same len g th .

1 07 . Leg en dre’

s-Th eoremmay be used for the approximat e

solution of sph erical trian g les in the followin g man n er .

( 1) Suppose t h e th ree sides of a spherical trian gle kn own ,

then th e values of a, B, y are kn own

,an d by t h e formulas of

Plan e Trig on ometry we can calculate S an d A’

,B

'

, C’

; th en

A,B, 0 ar e okn own

w

from the formqI l

Sr

(2) Suppose two sides an d t h e in cluded an gle of a spherical

t rian gle kn own,for example

L

A,b,c. Then

S 5By sin A’z: 5By sin A approximately.

Then A'

is kn own from t h e formula A Th us in the3r

2

plan e trian g le two sides an d the in cluded an g le ar e kn own ;therefore it s remain in g parts can be cal culat ed

,an d th en those

of t h e sph erical trian gle become kn own .

ON CERTAIN '

APPROXIMATE ' FORMULAE . ’

7 9

S uppose two sides an d t h e an gle opposite t o on e Of th emin a sph erical trian g le kn own , for example A,

a,b. Th en

3 Bsin B

’= — smA’

a. 0.

sin A approm'

mat e ly

an d O"

7 1 A’

B’= 7 r A B’

approximately ; the n S § aBsinHen ce A’

is kn own an d the plan e trian g le can be solved,sin ce two

sides an d the an g le Opposite to on e of th em are kn own .

(4) Suppose two an gles an d th e in cluded s ide of a sph e ri

cal trian gle kn own,for

“

example,A

,B,0.

Th en S — 72sin A

’

sin B’

f sin A sin B

Hen ce in t h e plan e trian gle two . an g le s an d t h e n in clude d side arekn own .

(5) Suppose two an gles an d t h e side opposite t o on e of th emin a sph erical trian gle kn own

,for example A

, B,a . Th en

7 : A’

B’

7 : A an d

a.

”sin B

'

sin 0’

2 sin (B’

which can be calculated, 81 11 063 . B

’

an d C”ar e approximately

kn own .

108. Th e importan ce of Legen dr e’s Th eorem in t h e application of Spherical Tri g on ometry to th e measur emen t Of t h e E arth ’ssurface h as g iven rise to various developmen ts Of it wh ich en ableus to test t h e deg ree of exactn ess of the approx imation .

We shallfin ish the presen t Chapter with some Of th ese developmen ts

,wh ich

will serve as exercises for the studen t. We have seen that ap

Sprox imately t h e sph erical excess is

'

equal ’ to an d we sh all

beg in wi th in vestigatin g a closer approximate formula for t h esph erical excess.

80 ON CERTAIN APPROXIMATE FORMUL E .

109. Tofin d an approximate value of th e sp h e rical e xce ss.

Le t E den ote the sph e rical excess 3 th en

sin § a sin § b sin 0 .

cos g.0

therefore approximately

sm § E sin 02262

GBF

GBth erefore E sm C5 ?

sin 0 ==sm

sm C'

cos 0’

aB3 2 1“

sm C 1

From (1)an d (2)

E : sin U’

é

GB

Hen ce to th is order of approximation th e area of t h e sph e ri

cal trian g le excee ds th at of t h e plan e trian gle by t h e fraction( 12+ B’ 7

2

2 473

of th e latter .

1 1 0. Tofin d an approximate value of

Sin A sin a.

Sin B si n b

sin Ah en ce approximate ly

sin B

ON CERTAIN APPROXIMATE FORMULZE . 8 1

182

64

B2

B4

B2

<B2

a2

)\

1 2W 3 6r4 l

B?

6?

6r"

7B2

1 1 1 . To expr e ss cot B cot A app r oxima te ly .

Cot B cot A1(cos B cos A);

sin B

h en ce, approximately, by Art . 1 10,

41 Ba

g —Bcot B — cot A

si n B (COS B _

6r2

Now we h ave sh ewn in Art . 1 06,t hat approx imately

B2

72

a.

2a4

B4

74 25

2

72

2 137

t h erefore cos B cosA approximately,

2 2 2 2 8

cot B cot AB B B 7 a

ay sin b’

ay sin B 1 2 r‘

2a

1 1 2 . Th e approximation s in Arts. 1 09 an d 1 10 ar e true so

far as terms in volvin g r"

; that in Art . 1 1 1 is true so far as

t erms in volvin g r”

,an d it will be seen that we ar e thus able

to carry the approximation s in t h e followin g Article so far as

t erms in volvin g r“.

T . S. T.

8 2 ON CERTAIN APPROXIMATE FORMULJE .

1 1 3 . To fin d an app roxima te valu e of th e e rror in th e le n g th

of a sid e of a sp h e rical tr ian g le wh e n calcu lated by Le g e n dr e’

s

Th e or em.

Suppose the side B kn own an d the side a required ; le t 3p. de

n ote the sph erical excess which is adopt ed. Then t h e approximate

value B s1 n (A _

’L)is taken for t h e side Of which a is the real

sin (B p.)

value. Le t x a W e h ave th en t o fin d x ap

proximately. Now approx imately

sin A —p cos A —’g

sin B p. cos B

l — pr cot A l — p co t B — fi2

sin A“

sin B (cot B — cot A)( 1 +p cot B).

Also the followin g formulae are true so far as terms in volv

iug r’

2 2 2

cot B— cot Aa

l + p cot B = 1 +

Hen ce,approx imately

,

sin A

sin B (cot B — cotA)( I p cot B)

84 MISCELLANEOUS EXAMPLES.

5 . I n a sph erical trian g le if A B 20,shew th at

a

cos a cos2

6 . ABC is a Spherical trian g le each of whose sides is a quad

ran t P is an y po in t with in t h e trian g le shew that

cos PA cos PB cosPC cot BPO cot CPA cot APB 0,

tan ABP t an BC'P tan CAP 1 .

7 . If 0 be t h e middle poin t of an equi lateral trian gle ABC,

an d P an y poin t on the surface of the sphere, then

fltan P0 tan 0A)2

(cos PA cos PB cosPC)“

cosgPA+cos

iPB+cos

’PC—OOSPAcosPB cosPB cosPO cosPC cosPA

8 . If ABC be a trian g le h avin g each side a quadran t, 0 t h e

pole of t h e in scribed circle,P an y poin t on the sph ere, th en

(008 PA cos PB cos 3 cos’PO.

9. From each of three poin ts on the surfac e of a Sph ere arcs

ar e drawn on t h e surface to thr ee other poin t s situated on a g reat

circle of the sphere, an d their cosin es ar e a,b, c ; a

’

,b’

,c’ ; a

”

,b”

,c”

Shew t h at ab”

c’

a’

bc"

a"

b’

c’

c”

a’

b"

c a”

bc

10. From Art s . 1 10 an d 1 1 1,sh ew that approx imately

log B= log a + log sin B- log sin A + - cot B).

1 1 . By con tin uin g t h e approximation in Art . 106 so as toin clude the term s in volvin g r

“

,shew that approx imately

By sinsA

’

By (a?

313237

2

)singA

'

cos A= cos A6r

21 80r

4

1 2 . From t h e precedin g result shew that if A A’

9 then

approximat ely

7 y9

X . GE ODE TICAL OPE RATIONS.

1 1 4. On e of the most importan t application s of Trig on o

metry,both Plan e an d Sph erical, is to t h e determ in ation of the

fig ure an d dimen sion s of the E ar th itself, an d of an y port ion of it s

surface. We Shall g ive a brief outlin e of the subj ect,an d for

further in formation refer t o Woodhouse’s Trig on ome try, to the

article Ge ode sy in the E n g lish Cyclop ae d ia, an d to Airy’s treatise

o n the Fig ur e of th e E ar th in the E n cyclop oe dia Me trop oli tan a .

For practical kn owledg e of the details of the operation s it will

be n ecessary to study some of the published accoun ts of the g reat

surveys which have been effected in differen t parts of t h e wor ld,as for example, the Accoun t of th e me asur eme n t of two se ction sqfth e Me r idion al arc of I n dia, by Lieut.-Colon el E verest, 1 847 or

the Accou n t of th e Obse rva tion s an d Calcu la tion s of th e Prin

cip al Trian g u la tion i n th e Ordn an ce Survey of Gr e at Bri tain

an d I r e lan d, 1 85 8 .

1 1 5 . An importan t part of an y survey con sists in t h e m e a

suremen t of a horizon tal lin e, which is called a base . A level plain

of a few mi les in len g th is selected an d a lin e is measured on it withevery precaution to en sure accuracy. Rods of deal

,an d of metal

,

h ollow tubes of g lass, an d st eel chain s,have been used in differen t

surveys ; the temperature is carefully observed dur in g t h e operat ion s, an d allowan ce is made for the varyin g len g ths of the rods

or chain s, wh i ch arise from variation s in the temperature.

1 1 6. At various poin ts of the coun try suitable station s ar e

selected an d Sig n als er e cted 5 then by supposin g lin es to be drawncon n ectin g the sig n als, the coun try is divided in to a series of

t rian g les. Th e an g les of these trian g les ar e observe d, that is, thean g les which an y two Sig n als subten d at a third. For example,suppose A an d B to den ote t h e extremities of the base

,an d 0 a

86 GEODETICAL OPERATIONS.

sig n al at a third poin t visible from A an d B then in ‘ t h e trian g leABC t h e an g les ABC an d BAO ar e Observed

,an d then AC an d BC

can be calculated. Ag ain , let D be a sig n al at a fourth poin t ,such that it is visible from C' an d A ; then the an g les AOB an d

CAD ar e observed,an d as AO is kn own ,

CD an d AD can be

calculat ed.

1 1 7 . Besides the orig in al base other lin es are measured in

con ven ien t part s of t h e coun t ry surveyed, an d their measured len g ths

ar e compared with their len g ths Obtain ed by calculation throug h a

series Of t rian g les from the orig in al base. Th e deg ree of closen ess with which t h e measured len g th ag rees with the calculated

len g th is a test of the accuracy Of the survey. Durin g the pro

g ress of the Ordn an ce Survey of Great Britain an d Irelan d,seve

ral lin es have been measured ; the last two are , on e n ear Loug hFoyle in Irelan d

,which was measured in 1 82 7 an d 1 82 8

,an d on e

on Salisbury Plain,which was measured in 1 849 . Th e lin e n ear

Loug h Foyle is n early 8 miles lon g , an d the lin e on Salisbury

Plain is n ear ly 7 miles lon g an d t h e differen ce between the len g th

of the lin e o n Salisbury Plain as measured an d as calculated from

the Loug h Foyle base is less than 5 in ches (An Accoun t of th e

1 1 8. Th ere are differen t methods Of effectin g t h e calculation s

for determin in g t h e len g ths of t h e Sides Of all t h e trian g les in t h e

survey. On e method is t o use the exac t formulae Of SphericalTr ig on ometry. Th e radius Of t h e E arth may be con sidered kn own

very approximately let this radius be den oted by r,t hen if a be

the len g th of an y arc t h e cir cular measure of the an g le which t h e

Th e formulae Ofa.

r

Spherical Trig on ometry g ive expression s for the trig on ometrical

arc subten ds at the cen tre of th e earth is

fun ction s Of 2, so th at 3may be foun d an d then a. Sin ce i n

a

practice is always very small,it becomes n ecessary t o pay

GEODETICAL OPERATIONS . 87

atten tion to t h e methods Of securin g accuracy in calculation swhich in volve the log arithmic trig on ometr ical fun ction s of small

an g les (Plan e Tr ig on ome try, Ar t .

In stead Of the exact calculation of the tr ian g les by Sph ericalTrig on omet ry, various methods of approximation h ave been pro

posed on ly two Of these methods however have been much used .

On e method Of approx imation con sists in deducin g from the an g lesof t h e spher ical tr ian g les the an g les of the ch ordal trian g le s , an dthen computin g the latter tr ian g les by Plan e Trig on ometry (se eAr t . Th e other method Of approximation con sist s in the

use of Leg en dre’

s Theorem (se e Ar t .

1 1 9 . Th e three methods which we have in dicated were al1

used by Delambre in calculatin g the trian g les in t h e Fren ch

survey (Base du Systems Mé tr iqu e , Tome I I I . pag e I n the

ear lier Operation s Of the Trig on ometrical survey Of Great Britain

an d Irelan d,t h e trian g les were calculated by the chord method ;

but this h as been for man y years discon tin ued,an d in place Of it

Leg en dre’

s Theorem h as been un iversally adopted (An Accoun t ofth e Obse rvation s pag e Th e t rian g les in the In dianSurvey ar e stated by Lieut. -Colon el E verest to be computed on

Leg en dr e’

s Th eorem . (An Accou n t of th e Me asur eme n t page

CLVI I I .)

1 20. If t h e th ree an g les of a plan e trian g le be Observed , t h efact that their sum oug ht t o be equal to two rig ht an g les affords a

test of the accuracy with which the Observation s ar e made. We

shall proceed to shew how a test Of the accuracy Of Observation s Of

the an g les of a spherical t rian g le formed on the E arth ’s sur face

may be Obtain ed by mean s of the sp h e rical exce ss.

1 2 1 . Th e ar e a of a sp h e rica l trian g le forme d on th e E ar th’s

surface be in g kn own in squar e f e e t , i t is r equir e d to e stablish a rule

for compu tin g th e sp h e rical exce ss i n se con ds.

Le t n be the n umber Of secon ds in the Spherical excess, s t h en umber of square feet in t h e area of the trian g le, r the n umber of


feet in t h e radius of the E arth . Then if E be t h e circular me asure of t h e spherical excess

,

s z E r f,

71 7 7 n

t 11 80 . 60 . 60 2062 65

appmmma e y

n r2

th erefore2 06 2 65

Now by actual measur emen t the mean len g th of a degreeth e E arth ’

s surface is foun d to be 365 1 5 5 feet ; th us

1 80365 1 5 5 .

With t h e value Of r Obtain ed from this equation it is foun d bylog ar ithmic calculation ,

that

log n log 3 9 3 2 67 7 4.

Hen ce n is kn own wh en 3 is kn own .

This formula is called Gen eral Roy’s rul e,as it was used by

h im in the Tr ig on ometrical survey Of Great Britain an d Irelan d .

Mr Davies, however , claims it for Mr Dalby. (Se e Hutton’

s

Course of Ma th emat ics,by Davies

,Vol . I I . p . 47 )

1 2 2 . I n Order to apply Gen eral Roy’s rule

,we must kn ow

t h e area Of the spherical t rian g le. Now t h e area is n ot kn ownexactly un less t h e elemen ts of t h e spherical trian g le ar e kn ownexactly ; but it is foun d that in such cases as occur in practice anapproximate value Of the area is sufficien t. Suppose

,for example,

that we us e the area Of the p la n e trian g le con sidered in Leg en dre’

s

Theorem,in stead Of the area Of t h e Sph e rical Tr ian g le i tse lf

then it appe ars from Art . 1 09,that th e error is approximately

B” y”

fraction is less than '0001,if the sides do n ot exceed 100 miles

in len g th . Or ag ain ,suppose we wan t to estimate the in fluen ce

of errors in t h e an gles on the calculation Of the area ; let t h e

den oted by the fraction of the former area,an d th is

G E ODETICAL OPERATIONS. 89

circular measure Of an error be h , so that in stead of

(0 h )2

approx imately the ratio expressed by h cot 0. Now in modern

Observation s h will n ot exceed the circular measure Of a few

secon ds,so that, if 0 be n ot very small

,h cot C’ is practically in

sen sible.

we Oug h t to usesm

t h e error th en bears to t h e area

1 2 3 . Th e followin g example was selected by Woodh ouse from

the tri an g les of the E n g lish survey,an d h as been adopted by other

wr iter s. Th e Observed an g les Of a trian g le bein g respectively

32 39 48 the sum Of the err0 1s made

in the Observation s is required, supposin g the Side Opposite t o thean g le A t o be 2 7 4042 feet. Th e area is calculated from the e x

a2sin B Sin O

2 sin A

that Now the sum Of the Observed an g les is 1 80°— 1

an d as it oug ht to have been it follows that the sumOf the errors Of t h e Observation s is Th is total error may

be dist ributed amon g the Observed an g les in such proportion as

the Opin ion Of the Observer may sug g est ; on e way is to in crease

each Of the Observed an g les by on e -third of an d t ake t h ean gles thus corr ected for the true an gles.

pression an d by Gen eral Roy’

s rule it is foun d

1 2 4. An in vestig ation h as been made with respect to t h e

form of a triang le, in which errors in the observation s Of the

an g les will exercise the least in fluen ce on t h e len g ths‘Of the Sides

,

an d althoug h the reason in g is allowed to be vag ue it may be

deservin g Of the atten tion of the studen t. Suppose the three

an g les Of a trian g le Observed, an d on e side,as a ,

kn own,it is

required to fin d the form Of t h e trian g le in order that the othersides may be least affected by errors in the Observation s . Th e

spherical excess Of the trian g le may be supposed kn own with

sufficien t accuracy for practice,an d if the sum Of the Observed

an g les does n ot exceed two rig ht an g les by the proper sphericalexcess, le t th ese an gles be altered by addin g t h e same quan tity to

90 GEODETICAL OPERATIONS .

each,so as to make their sum .correct. Le t A

,B

, 0 be ~the an g lest hus fur n ished by Observation an d altered if n ecessary ; an d le t

8A,8B an d 80 den ot e the respective errors Of A,

B an d 0 . Then

8A 8B + 80 0, because by supposition t h e sum Of A,B an d 0

is correct. Con siden the trian g le as approximately plan e, the

0 4- 80) a sin (0 + 80)t f ha s1 n (

ru e value 0 t e s1de 0 1s

si n (A + 8A)t h at i s

,sin (A — 8B — 80)

Now approximately

sin (0 80) sin 0 80 cos 0, (Pla n e Trig . Chap.

sin (A— 8B 80) sin A (8B 80)cos A.

Hen ce approximately

a sin 0sin A

1 + 80 00t 0 1

a

.

sm 01 + 8B cot A 80 (cot 0 + cot A)

S i n A

sin (A + 0) sin Ban d cot 0 + cot A

sin A sin 0 sin A sin Capproximately.

Hen ce the error Of c is approximately

a sin 0 cos A

singA

Similar ly t h e error Of b is approximately

a sin 0 a sin B cos A

singA

Now it is impossible to assig n exactly the sig n s an d mag n itudes

of t h e errors 8B an d 80,SO that the reason i n g must be vag ue. It

is Obvious that to make t h e error small sin A must n ot be small.

A n d as t h e sum Of 8A,8B an d 80 is zero, two Of them must have

the same sig n ,an d the third the opposite Sig n we may therefore

con sider that it is more probable than an y two as 8B an d 80 have

di fferen t sig n s, than that they ,

have the same Sig n .


small an g ular distan ce from t h e h orizon,the an g le wh ich the

Obj ects subten d bein g kn own , an d also t h e an gles Of e levationor depression .

Suppose OA an d OB the direction s in wh i ch the two poin ts

ar e seen from O an d let the an g le AOB be Observed . Le t OZ be

the direction at rig ht an g les to t h e Observer ’s horizon ; describe

a sphere roun d 0 as a cen tre,an d let vertical plan es throug h OA

an d OB meet the horizon at O0 an d OD respectively : then the

an g le 0OD is required.

Le t AOB = 0,

0 x,A00 = h

,BOD 75 ; from t h e

trian g le AZ B

cOS 0 cos Z A cos Z B cosrfi sin h sin h

COS AZ Bsin Z A sin Z B cos h cos h

an d cos AZ B cos 0OD cos (0 as) thus

cos O— sin h sin hCOS (6 + 96)

cos h cos it

This formula is exact by approximation we Obtain

cos d— x sin

ON SMALL VARIATIONS .

therefore a: sin 9 h h % (h2h f)cos 9, n ear ly,

2h le (h2+ h

e

)(cos2

5 9 sin"9)

2 Sin 9

i aw ym i e - I (h

Th l S process , by wh i ch we fin d the an g le 0OD from t h e

AOB,is called r e ducin g an an g le to th e h oriz on .

X I . ON SMALL VARIATIONS I N THE PARTS OF A

SPHE RICAL TRIANGLE .

1 2 8. It is sometimes importan t t o kn ow wh at amoun t Of

error w il l be in t roduced in t o o n e Of the calculated parts of a

t rian g le by reason of an y small error which may exist in the

g iven parts. We will here con sider an example.

1 2 9. A side an d th e Opp osi te an g le of a sp h e r ical trian g le

r ema in con stan t : de t e rmin e th e con n exion be twe e n th e small var ia

tion s of an y o th e r p air of e leme n ts.

Suppose 0 an d c t o remain con stan t.

( 1) Required the con n exion between the small variation s Of

the other sides . We suppose a an d b to den ote the sides Of on e

trian g le which can be formed with 0 an d c as fixed elemen ts, an d

a 8a an d b 8b to den ote the sides of an other such trian g le ;then we require the ratio of 8a to 8b when both ar e extremely

small. We have

cos c = cos a cos b+ sin a sin bcos 0,

cos c cos (a 8a)cos (b 8b) sin (a 8a)Sin (b 8b)cos 0 5

cos (a 8a) cos a sin a 8a,n ear ly

,

sin (a 8a) sin a cos a 8a, n early,

94 ON SMALL VARIATIONS IN THE

with similar formulae for cos (b 8b) an d sin (b (Se e Plan eTr ig o n ome try, Chap. XII .) Thus

cos 0 (cos a sin a 8a)(cos b sin 6 8b)

(sin a cos a 8a)(sin b cos b8b)cos 0.

Hen ce by subtraction,if we n eg lect the product 8a 8b,

0= 8a (sin a cos b— cos a sin c S O)

+ 8b (sin b cos a

this g ives t h e ratio of 8a to 8b in terms Of a, b, 0. We may

e Xpre ss the ratio more simply in terms Of A an d B ; for, dividin g

by sin a sin b, we g e t from Art . 44,

cot B sin 0 + .

8bco t A sin

sm b

th erefore 8a cos B 86 cos A z 0.

( 2 ) Required the con n exion between the small .variation s Of

the other an g les. I n this case we may by mean s of t h e polar

tr ian g le deduce from the r esult just foun d, t h at

8A cos b+ 8B cOs a—O;

th is may also be foun d in depen den t ly as before.

(3) Required t h e con n exion between the small variation s Of

a side an d the Opposit e an g le (A,a).

Here sin A sin c sin 0 sin a,

Sin (A + 8A)sin c = sin 0 8 in

hen ce by subtraction

cos A sin 0 8A sin 0 cos a 8a,

an d therefore 8A cot A 8a cot a.

(4) Required t h e con n ex ion between the small var iation s ofa side an d the adjacen t an g le (a, B).

96 E xAMPLE S .

5 . Supposin g B an d 0 to remain con stan t,prove t h e follow

in g equation s con n ectin g t h e small variation s Of pai rs Of t h e

other elemen t s

8b tan c = 8c t an b, 8A co t c = 8b sin A,

8A = 8a sin b sin 0, 8a sin B cos c = 8bsin A.

6. If A an d 0 ar e con stan t,an d b be in creased by a small

quan tity, shew that a will be in creased or dimin ished accordin g as

c is less or greater th an a quadran t .

X I I . ON THE CONNE X ION OF FORMULZE I N

PLANE AND SPHE RICAL TRIGONOME TRY .

1 3 1 . Th e studen t must h ave perceived th at man y of theresults obtain ed in Sp h e r ical Trig on ometry resemble others with

wh ich h e is familiar in l n e Trig on ometry. We Shall n ow pay

some atten t ion to this resemblan ce. We shall first shew h ow wemay deduce formulae in Plan e Trig on omet ry from formulae in

Spher ical Trig on ometry ; an d we shall then in vestig at e some

theorems in Spherical Trig on ometry which ar e in t erestin g prin cipally On accoun t of their con n exion with kn own results in Plan e

Geometry an d Trig on ometry.

1 32 . From an y formu la i n Sph e r ical Tr ig on ome try in volvin gth e e leme n ts of a tr ian g le , on e of th em be in g a sid e

,i t is r equir ed

to de duce th e corr e spon din g formula in Pla n e Tr ig on ome try.

Le t a, B, y be the len g ths Of the sides of the t rian g le, r the

a B r

7'

7"

7°

radius of t h e sphere,so that are t h e circular measures

of the sides of the trian g le ; expan d t h e fun ction s e f g,a.

which occur in an y proposed formula in powers ofr’

r e spe ct ive lv then if we suppose r to become in defin it ely g reat ,

CONNEXION OF FORMULA} IN TRIGONOMETRY. 97

t h e lim itin g form Of t h e proposed formula will be a relation

Plan e Trig on ometry .

For example, in Ar t . 1 06, from t h e formula

cos a cos bcos ccosA

sm ba n c

w e deduce

a4+B

4+ y

4 — 2 a2

32 — 26

2

y2 — 2 7

2a2

2 137 2 497 7”

n ow suppose r t o become in fin ite th en ul timately

297

an d th is is t h e expression for t h e cosin e of t h e an gle Of a plan e

trian gle in terms of t h e sides.

Agai n,in Art . 1 1 0, from t h e formula

sin A sin a

sin B sin b

w e deduceSin B B 637

3

n ow suppose r to become in fin i te th en ultimately

Sin'

A a.

smB_

B’

th at is,in a plan e trian gle t h e sides ar e as t h e Sin es Of t h e

site an g les.

1 33. To fin d th e equa tion to a small circle of th e

Th e studen t can easily draw t h e required di ag ram .

Le t 0 be t h e pole of a small circle, S a fixed poin t on t h e

Sph ere, SX a fixed g reat circle of t h e sphere. Le t OS r: a,

OSK B then the position Of O is determin ed by mean s of th esean g ular co-ordin ates a an d B . Le t P be an y poin t on the cir cum

feren ce of the small circle,PS : 9

,PSX 95, so that 9 an d (I) are

T. S . T. H

98 CONNEXION OF FORMULA}

t h e an gular co-ordin ates of P. Le t OP= r . Th en from t h e

trian g le OSP

—B)this g ives a relation between t h e an gular co-ordin ates Of an y poin ton the ci rcumferen ce of t h e circle.

If t h e circle be a g reat circle th en r th us t h e equation

becomes—B)

It will be Observed th at th e an g ular co-ordin ates h ere used ar ean alog ous t o t h e la titude an d lon g itude wh ich serve to determin e

t h e position s Of plac es on th e E arth ’s surface 9 is th e comp leme n tof th e la titude an d e is t h e lon g i tude .

1 34. E quation (1) of th e precedin g Article may be written

9cos r os

s

5sin

”

9 9cos a cos

2+ 2 sin a sin —

2- cos —

2—cos (ct —B).

9Divide by cos

’ —

2an d rearran g e ; h en ce

tan g —g: (cos r + cos a)- 2 tan gSin a cos (qS—B)+ cos r — cos a = 0.

Le t tan -

(

3,l an d tan Q? den ote th e values of tan g foun d from2 2

th is quadratic equation th en by Alg ebra, Ch apter XXI I.

tan -QJ tan -

o—g

cos r — cos az t an

a

jrtan

a — r

.

2 2 cos r + cOS a 2

Th us t h e value Of t h e product tan 92

‘tan0

52 is in d ep e n de n t of gt

this result correspon ds to t h e well-kn own property Of a circle in

Plan e Ge ometry which is demon strated in E uclid I II. 36 0orollary.

1 35 . Le t th ree arcs 0A, OB, 00 meet at a poin t. From an y

poin t P in OB draw PM pe rpen dicular to OA,an d PN pe rpe n

dicular to 00. Th e studen t can easily draw th e required diag ram

CONNEXION OF FORMULA?)

Th e studen t should cori'

vin ce h imself by examin ation that the

resul t holds for all relative posit ion s of P,P,,

an d P2 ,when due

reg ard is paid to alg ebraical sig n s.

1 38 . Th e prin cipal use Of Art . 1 37 is to determin e wh eth erth ree given poin ts ar e on t h e same great circle ; an illustrationwill be g iven in Art . 1 46.

1 39. Th e arcs drawn fi om th e an g les of a sp h e rical trian g le

p erpe n dicular to th e opp osi te sid e s r e sp e ctive ly me e t at a p oin t .

Le t OF be perpen dicular t o AB. From F suppose ar cs drawn

pe rpen dicular t o 0B an d 0A respectively ; den ot e t h e former by

f an d th e latte r by 77. Th en,by Art . 1 35

,

sin 5 sin FOB

sin 1] sin FOA

But , by Art . 65,

cos B cos OF sin FOB,cosA cos OF sin FOA

sin 5 cos B cos B cos 0fthere ore

si n 77 cosA cosA cos 0

An d if from an y poin t in CF arcs ar e drawn perpen di cular to

0B an d 0A respectively,the ratio of t h e sin e Of the former pe rpe n

di cular to the sin e of t h e latter perpen dicular is equal to ii iby Art . 1 35 .

IN PLANE AND SPHERICAL TRIGONOMETRY. 1 01

I n like man n er suppose AD perpen di cular to B0 ; then if from

an y poin t in AD arcs ar e drawn perpen dicular to A0 an d AB

respectively,the ratio of th e Sin e Of the form er pe rpen dicular to

cos A cos 0the sm e of t h e latter perpen di cular 1 s equal to

co s A cos B

Le t 0F an d AD meet at P,an d from P le t perpen diculars be

drawn on the Sides a,b,c of the trian g le ; an d den ote these pe r

pe n diculars by ac, y, 2 respectively then we h ave sh ewn th at

s in a: cos B cos 0

sin y cosA cos O’

sin y cos A cos 0d ish isan a

S I n z cosA cosB’

hen ce it follows th at

sin a: cos B cOS C'

sin z cos B cOS A’

an d th is sh ews th at th e poin t P 18 on t h e are drawn from B pe r

pe n di cular t o AO.

Th us -the three perpen diculars meet at a poin t,an d th is poin t

is determin ed by th e relation s

sin x sin y sin zcosB cos 0 cos 0 cosA co sA cos B

1 40. I n t h e same man n er it may be sh ewn th at t h e arcs

drawn from the an g les of a spherical trian gle to the middle poin ts

of the Opposite sides meet at a poin t an d if from this po in t arcs

w, y, z ar e drawn perpen dicular to the sides a, b, 0 respectively,

sin as sin y Sin zsm B sin O

_

sin 0 sin A“

sin A Sin B'

1 41 . It is kn own in Plan e Geometry that a certain circle

touches t h e in scribed an d escribed circles Of an y trian g le this

circle is called t h e Nin e p oin ts circle se e App e n dix to E uclid,pages 3 1 7 , 3 1 8, an d Pla/n e Trig on ome try, Ch apter XXIV.

‘

1 02 OONNEXION or FORMULzE

We sh all n ow sh ew th at a small circle can always be deter

min ed On the sphere to touch the in scribed an d escribed circles Of

an y sph erical trian gle.

1 42 . Le t 0. den ote t h e distan ce from A Of the pole Of t h e

smal l circle in scribed within a spherical trian g le ABO. Suppose

that a small circle of an g ular radius p t ouches this in scribed circle

in tern ally let B be the distan ce from A Of the pole of th i s touchin g cir cle ; le t 7 be t h e an g le between arcs drawn from A to thepole Of t h e in scribed circle an d t h e pole of t h e touchin g circle

respec tively. Th en w e must h ave

cos (p— r) cos a COSB+ Sin a sin Bcos y

Suppose t h at this touch in g circle also touches extern ally t h eescribed circle of an g ular radius r t hen if ( 1

,den ote t h e distan ce

from A Of t h e pole of this escribed circle, we must h ave

cos (p r,) cos a

,cos B sin a

,sin B cos y

Similarly, if a,an d at

3den ote the distan ces from A Of t h e poles

of t h e other escribed circles, in order that t h e touchi n g circle may

touch th ese escribed circles extern ally, we must also h ave

OOS agcosB+ sm a

gsc os

cos aacOSB+ sin a

asin Bcos

We sh all sh ew that real values of p, B, an d 7 can be foun d to

satisfy th ese four equation s .

E limin ate cos y from (1)an d (2) th us

cosP(cos r sin a,

cos r sin a)+ sin p (sin r sin a,

sin r,sin a)

B (cos a sin a,

cos a,sin a)

Suppose th at t h e in scribed circle touch es AB at the distan ce m

from A,an d th at t h e escribed circle of an g ular radius r

,touch es

AB at t h e di stan ce m,from A. Th en ,

by Art . 65,

1 04 OONNEXI ON or FORMULE

~ a . bWe may suppose th at cos -

2I s n ot less th an cos -

gor cos

that we ar e sure Of a possible value Of cos B from (9)

It remain s t o Sh ew. th atwh en p an d B ar e th us determi n ed,all

t h e four fun damen tal equation s ar e satisfied.

It will be Observed th at, p an d B bein g con sidered kn own ,cos y can be foun d from (1 )or an d

'

sin y can be foun d from

(3) or we must therefore shew that ( 1) an d (2) g ive t h esame value for cos y, an d that (3) an d (4) g ive the sam e value

for sin y ; an d we must al so sh ew th at th ese values satisfy the

con dition cos”

y sin”

yz 1 .

From ( 1)we h aveOOS p sin r

t + ta sm cotsm a

O r n p co r Sc 08 7 ,

b cCOS p Sl n

-

Qcos (s — a)s1n s cos —

2- cos

§—Z

2

sin Bcos y ;

t h is re duce s to

A b ccOS p S I I I

Qsm (b+ c)cos b

- cos

aSI n Bcos y

2

an d it will be foun d th at (2) r e duces to th e same so th at ( l)an d

(2) g ive the same value for cos 7 .

I n like man n er it will be f oun d th at (3) an d (4) agree inreducin g to

cos p cos 5sin (c b)cos

sin Bsin y.

I N PLANE AND S

It on ly r emain s to sh ew th at'

t h

satisfied.

Put It for that is —T

2

put X for cot r { 1 70 cos (3 an d Y for cot r, { 1 h cos

Th en (1)an d (2)may be wri tten respectively th us

(X cos p sin p)sin g(Y cos p

— Sln p)81n %From (10)an d (1 1)by addition

(X + Y)sin g—cOS p : 2 Sin Bcos y ;

t h erefore 4 sin”

B cos”

y (X”Y

22X Y)sin

”1;cos”But from (10)an d ( I 1 )by subtraction

(X Y)OOS p =— 2 sin p ;t herefore (X

”Y”

)cos”

p 4 sin”

p 2XY cos”

p.

Substitute in (1 2) an d we Obtain

B cos”

y==(sin

”

p XY cos”

p)sin

Again,put

for cot r2 { 1 h cos (s an d Y

,for cot r

3 { l — h cos (3

Th en (3)an d (4)may be writte n r espectively thus

(X ,COS p

— s1 n p)cos ij—z sin Bsin y

Ar

( 17 , C ( HF P)COS ?— 8m fism y ( l o).

From ( 1 4)an d (1 5)by subtraction

(X ,- Y

,)cosfg—OOs 2 sin Bsin y,

1 06 OONNEXION or FORMULE

an d from ( 14) an d ( 1 5)by addition,

(X ,Y,)OOS p = 2 sin p,

wh en ce

sin”

B (sin2

p X,Y,cos

”

p)cos”

%1

Hen ce from ( 1 3)an d ( 1 6) it follows th at w e h ave to establisht h e relat ion

B sin”

p XY sin”

gBut sin

”B 1 cos”B sin

”

p cos”

p h”cos

”

p, SO th at t h e r e

lation r e duces t o

1 1s: XY sin

” X,Y,t e e

?

Now

A cot r cot r, { l — h cos s } { l — k cos (s — b)sin (s—c)

2 sin b sin c

{ l — h cos s } { 1 — h cos (s — a)}sin bsin c

Similarly X Y 0082-4 { l - h cos (s — h cos (s — c)}2 S Ia I n c

Subtract t h e'

lat t e r from t h e former then we Obtain

h

m {cos (s - b)+ cos (s — c)— cos s - cos (s—a)}

[02

+Sm bsin c

{cos s cos (s — a)— cos (s - b)cos (s

2 h cosb+ 0

t h a””8sin b sin c 2 2

S

°

b+ c +a b+ c — a_ cos

a + c — b a + b

2 2 2 2

4 sin -

Ii sin -

(E- cos -

b- cos

e

2 2 2 2 h ”t h at i s

sm b sm c sm bsm c

th at is 1 la” wh ich was t o be sh ewn .

CONNEXION OF FORMULA]

COS Sl n

2

From (3)we se e th at we may put

gl e e s2

. c

2 2

b .

- c

”L .tang

fi T-

T—g

cos — + cos — cos2 2 2

Similar formulas of cours e h old for t h e poin ts of in tersection Oft h e touch in g circle with th e other side s.

1 45 . Le t z den ote the perpen dicular from t h e pole Of t h etouch in g circle on AB ; then

sin z = sin Bsin

B — Oos + cosA

2 7 28m

But from (2)an d (3) of Art . 1 42 w e h ave

cos si npa b e

811 1 6 008 7 T— Sm -

2- 81n —

2—Sln

b c awh ere Z = sm (s —a)— cos s srn (s — a)cos -

2cos§se c

z,

s inasin

bsin

c

2 2

b c a

wh ere Z,

a n (s b) cos (8 — c)mu (8 b)cos 2cos -

2se c

2

IN PLANE AND SPHERICAL TRIGONOMETRY. 1 09

Th erefore

sin z sin” Z cos

”sin E

b—sin -ésin g2 2 2

Now Z sin”-g

8m (8 a)Sin (8

.

6) 8m (8 _ c)1 — cos s cos é cos -gse c

S I n bsm c 2 2

an d Z,cos

Sin s sin (s — a)sin (e -; b) b c

sin bsin c

1 — cos (s— c)cos -

2- cos -

2se c

2A

2Th erefore Z S I n —

2Z,cos

is equal t o t h e product of

Sin (s a)sin (s b)sin b sin c

sin (s —c)+ sin s — cos —b-cos —

c- sec

a

2 2 2

sm (s — a)Sin (s — b)sin b sin c

Sin (8 — a)sin (s b)

2 Sin bsin§

sin (s — a)sin (s — b)sin

0

2sin bsin

sin (8 a)sin (3 b)sin”

sin bsin6

cos“

2 2

(s — c)cos s + cos (s — c)sin s

b c acos —

2cos

2se c —

2sm (2s c)

b a—s1 n (a b)cos g

se c

?

J

Sin (8 a)sin (8 b)sin”

2 cOSa

cOSbsin

0a d

2 2 2

1 1 0 OONNEXION OF FORMULE

Th ere fore1

a b csm (s — a)s1 n (s — b) l

S I n —

2~ S I n —

2- sm

§ c .

1 }

sm”

§s1 n a sm b J

cOSp b c A — B

Tsinzsin

zsin

z2 cos

”

2

COSp . a b .

Tsingsin -

z—sin

2

ccos (A B)

z = sin p cos (A - B).

Similar expression s h old for t h e perpen diculars from the pole

t h e touch in g circle o n the oth er sides Of t h e sph erical trian gle .

1 46. Le t P den ote the poin t determ in ed in Ar t . 1 39 0 t h e

poin t determin ed in Art . 1 40, an d N the pole Of the touch in gcir cle. We shall n ow shew th at P,

0,an d N ar e on a g reat

Le t r, y, z den ote t h e perpen di culars from N on t h e sides

a,b,0 respectively Of t h e Spherical trian gle ; let y,, 2 , den ote

t h e perpen diculars from P ; an d x2 , y, , z , the perpen diculars from

0. Then by Ar ts. 1 45,1 39

,an d 1 40 we h ave

sin a: sin y sin

cos (B 0) cos (0 —A) cos (A - B)

S I n cc,

sin y , sin z,

cosB cos 0 cos 0 00s A cos A cos B’

sin as,

sin y, sin z2

sin B sin 0 sin 0 sin A sin A sin B'

Hen ce it follows that

sin ce : t,mum

,t2sin ac

, ,

S i n z t,sm z

,+ t

,SI n z

,,

1 1 2 EXAMPLES.

E XAMPLE S .

1 . From th e formula sin 900 ? S cos (S _ A) deduce

2 sm B S I n 0

th e expression for t h e area Of a plan e trian g le, n amely

a”sin B sin 0

2 sin A

creased.

wh en t h e radius of t h e Sph ere is in defin itely in

2 . Two trian g les ABO, abc, sph erical or plan e,equal in all

respects,differ slig htly in position sh ew th at

cosABbcosB00 cos 0Aa cosA0c cos OBbcos BAa O.

3 . Deduce formulas in Plan e Tr igon ometry from Napier’s

An alogies.

4. Deduce formulae in Plan e Trigon ometry from Delambre’s

An alogies.

c A +B 0 a + b5 . From t h e formula cos -

2- cos

2sm -

gcos

2deduce

the area of a plan e tri an g le in terms Of t h e sides an d on e of the

an g les.

6 . Wh at result is Obtain ed from E xample 7 to Ch apter VI .

,

by supposin g t h e radius Of t h e sph ere in fin ite ?

7 . From t h e an g le 0 of a sph erical trian g le a perpen di cular is

drawn to the ar e which j oin s the m iddle poin ts Of the sides a an d

b : Shew that this perpen dicular makes an an gle S —B with the

side a,an d an an g le S —A with the side b.

8. From each an g le Of a sph erical trian g le a perpen dicular I s

drawn to the arc which j oin s the mi ddl e poin ts Of the adjacen tsides. Sh ew th at these perpen diculars m e et at a poin t ; an d that

EXAMPLES. 1 1 3

if x, y, z ar e t h e perpen diculars from th is poin t on t h e sides a

,b, 0

r espectively,

Sin x Sin y sin zsin (S —B)sin (S 0) sin (S 0)sin (S —A) sin (S—A)Sin (S —B)

9. Th roug h each an g le Of a sph erical trian g le an arc is drawnso as to make the same an g le with o n e side wh ich t h e pe rpe n

dicular on t h e base makes with the other Side. Shew that th esear cs meet at a poin t ; an d that if x

, y, z ar e t h e perpen dicularsfrom th is poin t on t h e Sides a

,b,c respect ively

,

sin x sin y sin z

cosA cos B cos 0°

1 0. Sh ew th at t h e poin ts determin ed in E xamples 8 an d 9,

an d the poin t N of Ar t . 1 46 ar e on a g reat circle.

State the correspon din g th eorem in Plan e Geometry.

1 1 . If o n e an g le of a sph erical trian g le remain s con stan t wh ilet h e adjacen t sides ar e in creased

,sh ew th at t h e area an d t h e sum

of t h e an gles ar e in creased.

1 2 . I f t h e arcs bisectin g two an gles Of a Sph erical trian gle an dtermin ated at t h e Opposit e sides ar e equal , the bisected an g les wil lbe equal provided their sum be less th an

[Le t BOD an d COE den ote these two arcs wh i ch ar e givenequal. If the an g les B an d 0 ar e n ot equal Suppose B the greater .

Then is g reater than BE by Art . 5 8 . An d as t h e an g le OBOis greater than the an g le OOB,

th erefor e 00 is g r eater th an OB ;

therefor e OD is greater than OE . Hen ce t h e an gle OD0 is

g reater than the an g le OE B,by E xample 1 1 . Th en con struct

a spherical trian gle BOF on the other side Of B0, equal to OBE .

Sin ce the an gle is greater than t h e an g le OE B,the an g le

FDO is g reater th an the an g le DFO; therefore is less th an

OF ,so that 0D is less than BE . Se e t h e correspon din g problem

in Plan e Geometry in the App e n dix to E uclid , pag e

T. S. T.

( 1 1 4 )

X III. POLYHE DRONS.

1 49. A polyh edron is a solid boun ded by an y n umber Ofplan e rectilin eal fig ures wh ich ar e called it s faces. A polyhedronis said to be r eg ular when it s faces are similar an d equal regular

polyg on s, an d its solid an gles equal to on e an oth er.

1 50. If S be the n umbe r of solid an g le s in an y p olyh edron ,

F th e n umber qf i ts face s, E th e n umbe r of i ts edg es, th e n

Take an y poin t with in t h e polyh edron as cen tre,an d describe

a sph ere Of radius r,an d draw straig ht lin es from the cen tre

to each of t h e an g ular poin ts of t h e polyh edron ; le t t h e poin tsat wh ich th ese straig h t lin es meet the surface Of t h e sphere bejoin ed by arcs of g reat circles, so th at t h e surface of the sph ere isdivided in to as man y polygon s as the polyhe dron h as faces.

Le t 3 den ote th e sum of t h e an g les Of an y on e Of th ese poly

g on s, m t h e n umber of it s sides th en th e area of t h e polyg on isr”

{s— (m — 2)w} by Art . 99. Th e sum Of the areas Of all t h epolyg on s is th e surface Of t h e sphere

,that is

, 41rr”. Hen ce sin ce

t h e n umber of t h e polyg on s is F ,w e Obtain

41 r = E s — 7r2m+ 2 E rr.

Now E s den ote s t h e sum Of all the an gles Of t h e polyg on s, an dis therefore equal to 2 7 : x the n umber of solid an g le s, that is, t o2 7rS ; an d 2 m is equal to the n umber of all th e sides of all t h e

polyg on s, that is, to 2 E ,sin ce every edg e g ives ri se to an arc

which is common to two polyg on s. Th erefore

417 : 2 7rS —2 7 t E + 2 E rr ;

th erefore S F E 2 .

1 1 6 POLYHEDRONS.

demon strated th at, th e r e can n ot be mor e th an five solids e ach ofwh ich h as a ll i ts face s wi th th e sam e n umbe r qf sid e s

,an d all i ts

solid an g lesformed wi th th e same n umbe r ofp lan e an g le s .

1 5 2 . Th e sum of all th e p lan e an g le s wh ich form th e solid

a/n g le s of a/n y p olyh e dron is 2 (S 2)7r.

For if m den ote the n umber of sides i n an y face of t h e polyh e dro n

,t h e sum of the in t erior an g les Of that face is (m — 2)1 r

by E uclid I . 3 2 , Cor . 1 . Hen ce t h e sum Of all the in terior an g lesof all t h e faces is 2 (m — 2)7r, that is E ma 2F 7 r

,that is

2 (E — E )rr, that is 2 (S — 2)7r.

1 5 3. Tofin d th e in clin ation of two adjace n tface s of a regular

Le t AB be t h e edg e common t o t h e two adjacen t faces, 0 an d

D t h e cen tres Of the faces bisect AB at E,an d join OE an d DE ;

OE an d DE will be pe rpen dicular to AB,an d the an g le OE D is

the an g le of in clin ation of the two adjacen t faces we sh all den ote

it by I . I n the plan e con tain i n g OE an d DE draw 00 an d DO

at rig ht an g les to OE an d DE respectively, an d meetin g at 0 ;

about 0 as cen tre describe a sphere meetin g 0A, 00, OE at a,c, e

respectively, so that .cae forms a spherical trian g le. Sin ce AB is

perpen dicular to OE an d DE ,it is perpen dicular to the plan e

OE D,therefore the plan e AOB which con tain s AB is perpen dicular

t o the plan e OE D hen ce the an g le ce a of t h e spherical trian g le is

a rig ht an g le. Le t m be the n umber Of sides in each face Of the

polyh e dron , n th e n umber of the plan e an gles wh ich form each solid

POLYHEDRONS. 1 1 7

2Th en t h e an g le ace z AOE

fl

an d t h e an g le2m m

on e of t h e n equal an g les formed o n the sph ere roun d a,

2 71' 77

2 7, aFrom the righ t—an gled tr ian gle cae

cos cae cos cOe sin ace,

th at is _

1_

therefore

1 5 4. To fin d th e radi i of th e i n scr ibed an d circumscr i b edsp h e r e s of a r eg ular p olyh e dron .

Le t t h e edg e AB z a,le t 00 = r an d 0A=B,

so th at r is

t h e radius of the in s cribed sph ere,an d B is t h e radius Of t h e

circumscr ibed sph ere. Th en

0E = AE cot AOE =g

cot I ,2 m

r = OE tan 0E O= OE tan g cot z tanli

2 2 m 2

r = R cos aOc= R cot e ca cot e ac z c ot -ZZ cot —Z ;

th erefore B z r tan E - tani—rg tan —

I- tan

r

.

m n 2 2 n

1 5 5 . Tofin d th e surface an d volume of a re g ular p olyh e dron .

Th e area of on e face of th e polyh edron is £2? cots-72,2

th erefore t h e surface of t h e po lyh edron isml

ja

2;Al so t h e volume Of t h e pyrami d wh ich h as on e face of t h e

2

polyh edron for base an d O for vertex is 3.

”

if cot7

12, an d

mFra2

7 1

cotth erefore t h e volume of th e polyh edron is1 2 m

1 1 8 POLYHEDRONS .

1 5 6. To fin d th e volume of a paralle lep ip ed in t e rms of i ts

edg es an d th e ir in clin ation s to on e an oth e r .

Le t th e edges be OA= a, 0B =b

,00 = c le t t h e in clin ation s

be BOO a, COA B, AOB z

y. Draw OE perpen dicular to theplan e AOB meetin g it at E . Describe a sphere w ith O as a

cen tre,meetin g 0A,

OB,00, OE at a

,b, c, e respectively.

Th e volume of t h e parallelepiped is equal to t h e product of it s

base an d altitude absin y OE abc sin y sin cOe . Th e spher icaltrian gle cae is rig ht-an g led at c th us

sin c0e = sin cOa sin cae = sin Bsin cab,

an d from th e sph erical trian g le cab

2 2 2

sin cab J( l cos a cos B.

cos

.

7 + 2 cos a cOSBcos y);

81 11 B sm y

th erefore t h e volume of th e parallelepiped

abc J( 1 — cos”a cos

”

B 2 cos a cosBcos y).

1 5 7 . To fin d th e diag on al of a paralle lep ip ed in t e rms of th eth re e edg e s wh ich i t me e ts an d th e ir in clin at ion s to on e an oth e r .

Le t t h e edg es be 0A a, 0B b

, 00= c le t t h e in clin ation s

be BOO= a, OOA =B, AOB z

y. Le t 0D be the diag on al

quired,an d let OE be the di agon al of the face OAB. Th en

OD”: OE

” E D”+ 2OE . E D cos OOE

a” b

”+ 2abcos y + c

”+ 2 cOE cos OOE .

1 20 POLYHEDRONS.

1 44 V”

a’”b

’”c'”

a”a'”

(b’”

c'”

b”b’”

(c’”

a’”

b c”c'” (a b c

a'”

(a” b”)(a

” c”) b'”

(b” c”)(b

”a”

) c’”

(c”

a”

)(c”

Th us for a r e g ular tetrah edron we have 1 44 V”2 a

”.

1 5 9 . If the vertex Of a tetrahedron be supposed to be situ

ated at an y poin t in t h e plan e of it s base,the volume van i shes ;

h en ce if we equat e t o z ero the expression on th e rig h t-han d sideOf the equation jus t g iven , we Obtain a relation which must holdamon g t h e six straigh t lin es wh i ch j oin four poin t s taken arbi

t rarily in a plan e.

Or we may adopt Carn ot ’s meth od,in wh i ch th is relation is

established in depen den tly, an d t h e expression for the volume of a

tetrah edron is deduced from it ; th is we shall n ow shew,an d we

sh all add some oth er in vestig ation s wh ich ar e also given by

Carn ot.

It will be con ven ien t to alter the n otation h i th erto used,by

in terch an gin g the accen ted an d un accen ted letters.

1 60. To fin d th e r e la tion h oldi n g amon g th e six straig h t lin e s

wh ich joi n four p oin ts take n arbi trar ily i n a p lan e .

Le t A,B, 0, D be the four poin ts . Le t AB 0

,B0 a

,

OA b ; also let DA = a'

,DB : b

'

,DO c

'

.

If D falls w ith in t h e trian g le ABC,t h e sum of

i

t h e an g les

ADB,BDO, ODA is equal to four rig ht an g les so that

cos ADB cos (BDO ODA).

Hen ce by ordin ary tran sformation s we de duce

1 cos”ADB cos

”BDO cos

”ODA 2 cosADB cosBDOcos ODA.

If D falls wi thout t h e trian g le ABO, on e Of t h e three an g les

at D is equal to the sum Of t h e oth er two,an d t h e resul t just

given still holds.'2 1 2

Now cosADBa 0

9

an d t h e other cosin es may be

expressed in a similar man n er ; substitute th ese value s in t h e

POLYHEDRONS. 1 2 1

above result,an d we Obtain t h e required relation

,wh ich after

reduction may be exhibited th us,

0 a”b”c”

a’”a” c” a

”

) b’”b”

(c”

a”b”

) c'”c”(a

”b” c”)

a”

(a (a'” b

2

(b’”

(b'”

a c” (c'”

a (c b

1 6 1 . To exp r e ss th e volume of a t e trah e dron in t e rms of i ts six

edg es .

Le t a,b,c be the len gths Of t h e sides Of a trian g le ABO

formin g on e face Of the tetrahedron,wh ich we may call it s base ;

le t a’

,b'

,c'

be the len g th s Of the straig h t lin es which j oin A,B,0

respectively to the vertex of the tetrahedron . Le t p be t h e len g th

of the perpen di cular from the vertex on the base then the len g ths

Of the straig ht lin es drawn from t h e foot Of the perpen dicular to

A,B,0 r espec t ively ar e J(a J (b

'” Hen ce

t h e relation g iven in Ar t . 1 60 will hold if we put (a p”

) in

stead Of a'

, J (b’”

p”

) in stead of b ’,an d J(c

’”

p”

) in stead of c’.

We sh all th us Obtain

p”

(2a”bz2b 0 2 0

”a”

a4b” c”) a

”b”c”

a’”a”

(b” c” a

”

) b’”b”

(c”

a”b”

) c'”c” (a

” b” c”)

a”

(a (a’”

b”

(b'” c (b

'” c” (c'”

(c

Th e coefficien t of p”i n th i s equation is sixteen times t h e

square of t h e area of t h e trian gle ABO so th at the left -h an dmember is 1 44 V”, wh ere V den otes t h e volume of t h e t e t rah e

dron . Hen ce th e required expression is Obtain ed.

1 62 . To fin d th e r e la tion h oldin g amon g th e six arcs of g r e a tcircle s wh ich join four p oin ts tah e n arbi trarily on th e surf/

"

ate of a

sp h e r e .

Le t A, B, 0, D be t h e four poin ts. Le t AB z

y,‘

B0 a,

CA =B; let DA = a'

,DB z B

’

,D0 = y

’

.

As in Ar t . 1 60 we h ave

1 cos”ADB cOS

”BD0 cos

”ODA 2 cosADB cosBDO cosODA.

1 2 2 POLYHEDRONS .

cosy cos a’

cosB’

sin a’

sin B’

may be expressed in a similar man n er substitute th ese values int h e above result

,an d we Obtain the required relation ,

which afterreduction may be exh ibited thus,

Now cos ADB an d t h e oth er cosin es

l cos”a cos

”

B cos”

y cos”a’

cos”

B’

cos”

7’

cos”a cos

”a’

cos”

B cos”B

’

cos”

7 cos”

y’

2 (cos a cos B cOS y cos a cos B’cos y

’

cos B cos a’

cos y’

cos 7 cos a.’

cos B’

)

2 (cos (1 cos B cos a’

cos B’

cos B cos 7 cos B’

cos y’

cos 7 cos a cos y’

cos

To fin d th e radius of th e sp h e r e circumscribin g a te tra

Den ote t h e edg es Of t h e tetrahedron as in Art . 1 61 . Le t thesphere be supposed t o be circumscr ibed about the tetrah edron ,

an d draw on the sphere the six arcs Of g reat circles join in g thean g ular poin ts Of t h e te trahedron . Then the r elation g iven in

Art . 1 62 holds amon g t h e cosin es of these six ar cs.

Le t r den ote the radius of the sphere. Th en

a a”

°

s_cos a — 1 2 s1 n2 2 r

” ’

an d the ot h er cosin es may be expressed in a similar man n er .

Substitute these values in the result of Art . 1 62 , an d we Obtain ,

after reduction,with t h e aid Of Art . 1 6 1 ,

4 x 1 44 Vgr”

b”b’”

Th e rig h t -han d member may also be put in to factors , as we se e

by recollectin g t h e mode in ,wh ich t h e expression for the area Of

a trian gle is put in to factors . Le t aa’

bb’

cc’

then

36 V2?”

0'

(O'

d a’

)(0' bb

'

) (O'

1 24 EXAMPLES.

1 1 . If wit h all the an g ular poin ts Of an y paralle lépipe d as

cen t res equal spheres be described,the sum Of the in tercepted

portion s Of the parallelepiped will be equal in volume t o on e Oft h e spheres .

1 2 . A reg ular octah edr on is in scribed in a cube so that thecorn ers Of t h e octahedron ar e at the cen tres Of the faces Of thecube : sh ew that the volume Of t h e cube is six times th at of t h eoctah edron .

1 3 . It is n ot possible to fill an y g iven space with a n umber

of regular polyh edron s of the same ki n d,except cubes ; but this

may be don e by mean s of tetrah edron s an d octah edron s wh ich

have equal faces,by usin g twice as man y Of th e former as Of

t h e latter.

1 4. A spherical trian g le is formed on t h e surface Of a sph ere

Of radius p it s an g ular poin ts ar e join ed,formin g thus a pyramid

with t h e straig h t lin es j oin in g th em with the cen tre : sh ew th att h e volume Of t h e pyramid is

0

é—pa

J(tan r tau r,t an r

2tan

wh ere r,r,,r2 ,r3ar e t h e radii of t h e in scribe d an d escribed cir

cle s of t h e t rian gle.

1 5 . Th e an g ular poin ts Of a reg ular tetrah edr on in scribedin a sph ere of radius r bein g taken as poles, four equal smal l

circles of t h e sphere ar e described,so th at each circle touch es

t h e other three. Shew that t h e area Of t h e surface boun ded by

each circle is 2 1rr”

1 6 . If 0 be an y poin t with in a spherical trian g le ABO,t h e

product of the Sin es of an y two sides an d th e sin e of t h e in

cluded an g le

—sin A0 sin BO Sin 00 AO sin BOO

cot B0 sin COA cot 00 sin AOB

X I V . ARCS DRAWN TO FIX E D POINTS ON THE

SURFACE OF A SPHE RE .

1 64. I N t h e presen t Ch apter we sh all demon strate various

proposition s relatin g to the arcs drawn from an y poin t on t h e

surface of a sphere t o certain fixed poin ts on t h e surface.

1 65 . ABO is a sph erical tri an gle h avin g all it s sides quadran ts, an d th erefore all it s an gles rig h t an gles ; T is an y poin ton t h e surface of the sph ere : t o shew that

cos”TA cos

”TB cos

”TO 1 .

By Art . 37 we h ave

cos TA cosAB cos TB sin AB sin TB cos TBA

sin TB cos TBA.

Similarly cos TO sin TB cos TBO sin TB sin TBA.

Square an d add ; th us

cos”TA cos

”TO 2 sin

”TB 3 1 cos

”TB

th erefore cos”TA cos

”TB cos

”TO z 1 .

1 2 6 ARCS DRAWN To FIXED POINTS

1 66. ABO is a sph erical"

trian g le h avin g all its sides quadran ts, an d therefore all it s an g les rig h t an g les ; T an d U ar e an ypoin ts on t h e surface Of th e sph ere : to sh ew that

cos TU: cos TA cos UA co s TB cos UB + cos TOcos 00.

By Art . 37 we h ave

cos TU= cos TA cos UA+ sin TA sin UA cos TAU,

cos TAU: cos (BAU—BAT)

cos BAO cosBAT + sin BAUsin BAT

cos BAUcosBAT + cos OAUcos OAT ;

th erefore cos TU cos TA cos UA

+ sin TA sin UA (cosBAUcosBAT + cos OAUcos OAT);

cos TB sin TA COS BAT,

cos UB sin UA cosBAU,

cos TO sin TA cos OAT,

cos UO==sin UA cos OAU;th erefore

cos TU: cos TA cos UA cos TB cos UB + cos TO cos 00.

1 67 . We leave to t h e studen t t h e exercise Of sh ewin g th att h e formulas of t h e two precedin g Articles are perfectly g en eral for

all position s of T an d U,outside or in side t h e trian gle ABO t h e

1 28 ARCS DRAWN TO FIXED POINTS

1 70. We sh all n ow tran sform the resul t Of Art . 1 68.

Le t G AP” Q”

an d let a, B, y be th ree arcs determin ed by the equation s

P Q Bcos a

a: B "

G,

( 308 7“

at ;

Sin ce cos”a cos”B cos”y 1

,it is Obvious th at there will be

some poin t o n the surface of the sphere,such that a

, B, y ar e thearcs which j oin it to A

,B, 0 respectively ; den ote th is poin t by

17 : th en ,by Art . 1 66

,

cos TU z h cos a + pt cosB+ v cos y ;

an d fin ally2 G TUcos

Th us, wh atever may be the position Of T,t h e sum of t h e cosin es

Of the arcs wh ich join T to the fixed poin ts varies as the cosin e

Of the sin g le arc wh i ch join s T to a certain fixed poin t 17 .

We mig h t take 0 eith er positive or n eg ative ; i t will becon ven ien t to suppose it positive .

1 7 1 . A sph ere is described about a reg ular polyh edron

from an y poin t on the surface Of t h e sph ere arcs ar e drawn to t h e

solid an g les Of t h e polyhedron t o shew that t h e sum Of th e cosin es

of these arcs is zero .

From the precedin g Ar ticle we se e th at if 0 is n ot z ero

there is on e posit ion of T which g ives to 2 it s g reatest positivevalue

,n amely, when T coin cides with U. But by the symmetry

Of a reg ular polyhedron there must always be mor e th an on e posi

tion Of T which g ives t h e same value t o E . For in stan ce, if we

take a reg ul ar tetrahedron ,as th ere ar e four faces there will at

least be th r e e oth e r position s Of T symmetrical with an y assign ed

position .

Hen ce 0 must be zero ; an d th us t h e sum of the cosin e s of th e

arcs wh ich join T to th e solid an g le s Q ] th e r eg ular p olyh e dron is

z e rofor a ll p osi tion s of T.

ON THE SURFACE OF A SPHERE 1 2 9

1 7 2 . Sin ce 0 : 0, it follows that P, Q, I t must each be zeroth ese in deed are particular cases of the gen eral result Of Art . 1 7 1 .

Se e Ar t . 1 69.

1 7 3. Th e result Obtain ed in Art . 1 7 1 may be shewn to hold

also in some other cases . Suppose,for in stan ce

,that a r e ctan g u

lar parallelepiped is in scribed in a Sphere then the sum Of the

cosin es of the arcs drawn from an y poin t on the surface of the

sph ere to t h e solid an g les Of the parallelepiped is zero. For her e

it is Obvious that there must always be at least on e oth e r position

of T symmetrical with an y assig n ed position . Hen ce by t h e

argumen t Of Art . 1 7 1 we must have 0 z 0.

1 74. Le t there be an y n umber Of fixed poin ts on the surface

Of a Sph ere den ote them by HpI I

Q ,H

3, Le t T be an y poin t on

the surface of the Sphere. We shall n ow in vestig ate a remarkable

expression for the sum Of the squares of t h e cosin es Of t h e arcswhich j oin T with the fixed poin ts.

Den ote t h e sum by E so that

2 = cos”TH

,cos

”TE

2+ cos

”TE

3

Take on t h e surface Of t h e sphere a fix ed spherical trian gleABO, havin g all it s sides quadran ts

,an d th erefore all it s an gles

rig h t an gles.

Le t A, p , v be t h e cosin es Of t h e arcs wh ich join T with A,

B, 0

respectively ; let l,, m,,n,be t h e cosin es Of t h e an g les w hich j oin

H,with A, B, 0 respectively ; an d le t a similar n otation be used

with respect to H2 ,

Then,by Art . 1 66,

2 mm. n,v)”

(lzk mg r-t n

zv)”

E xpan d each square,an d rearran g e t h e terms th us

2 2 PA.

”

Opt”RV

22ppw 2 g rh .

wh ere P stan ds for l,

”l,

”

an d p stan ds for m n m,,n,m

,,ug

with correspon din g mean in gs for Q an dq, an d for B an d r .

T. S . T.

1 30 A RCS DRAWN TO FIXED POINTS

We Sh all n ow sh ew th at there is some position of t h e t r ian g leABO for wh ich p , q, an d r will van ish ; so th at w e sh all thenhave

2 Pr or as .

Sin ce 2 is always a fin i te positive quan tity there must be some

posit ion , or some position s , of T for which 2 h as the larg est value

which it can r eceive. Suppose that A h as this position , or o n e

Of these position s if there ar e more th an on e . When T is at A

we have p. an d v each zero, an d A equal t o un ity, so that 2 is then

equal to P.

Hen ce,whatever be t h e position Of T,

P is n ever less th an PA” Op.”By

”21)/w 2 g yk 2 r)\p ,

th at is,by Art . 1 65 ,

P (A?

p.”

v”

) is n ever less th an

PA” Op.”Bv

” 2pwr 2qv>t 2 rApth erefore

(P Q)p.”

(P R)v” is n ever less th an 2ppa+ 2qv} t +

Now suppose v= O ; then T is situated on t h e g reat circle Ofwhich AB is a quadran t, an d wh atever be t h e position of T w e

have

(P O)p.

”n ot less than

an d therefore P Q n ot less th an2

7

7

1?But n ow é is e ual t O

COS TA° t h is

'

n°

all n al toM

qcos TB

I s ume rI c y eqtan TB

,an d so may be made n umerically as g reat as we please

,

positive or n eg ative, by givin g a suitable position to T. Thus

P 0must in some cases be less th an if r h ave an y value dif~IL

fe r e n t from zero.

Therefore r must 0.

I n like man n er we can sh ew th atqmust 0.

1 3 2 ARCS DRAWN TO FIXED POINTS

With t h e n otation of Art . 1 7 4 we h ave

2 PA”

0p.”By

”.

We shall shew that in the presen t case P, Q, an d B must allbe equal. For if they ar e n ot

,on e Of them must be g reat er than

each Of the others,or on e Of them must be less than each Of the

others .

If possible let the form er be the case ; suppose th at P is

g reater than 0, an d g reater than R.

2 =P (1 — p.

”

— P - m— ve

this sh ews th at 2 is always le ss th an P except when pt O an d

v O th at is E is a lways le ss th a n P except wh en T is at A, or

at the poin t Of the surface wh i ch is diametrically Opposite t o A.

But by the symmetry Of a reg ular polyhedron there must alwaysbe more than two position s Of T which g ive the same value to E .

For in stan ce if we take a reg ul ar tetrahedron ,as there ar e four

faces there will be at least th r e e oth e r position s Of T symmetrica l

t h an y assig n ed position . Hen ce P can n ot be greater than Qd g reater th an R.

I n the same way we can shew th at on e Of t h e three

an d B,can n ot be less than each of the others.

Therefore P 0 B an d th erefore by Art . 1 65 for every

p osit ion of T we have 2 P.

Sin ce P Q B each Of them

1

3 {l,

”m

,

”n,

”l,

”m,

”n ,”

by Art . 1 65,

where S is t h e n umber of the solid an gles Of th e reg ular polyh e dron .

ON THE SURFACE OF A SPHERE. 1 33

Th us th e sum of th e squar e s of th e cosin e s of th e arcs wh ich

join an y poin t on t h e surface of th e sp h e r e to th e solid an g le s ofth e r eg ular p olyh e dron is on e th ird of th e n umbe r of th e solid

an g le s .

1 7 8 . Sin ce P : Q R in t h e precedin g Article,it wil l follow

that when t h e fixed poin ts Of Ar t . 1 7 4 ar e the solid an g les Of

a r eg ular polyhedron ,then for an y position Of the spherical t ri

an g le ABO we shall have p 0,qO

,an d r = O.

For takin g an y position for the Spherical trian gle ABO we

h ave2 PA

”

Q B”RV

”2pm! 2qvh

then at A we h ave pt : 0 an d v : 0,SO th at P is then t h e val ue

of E ; similar ly Q an d B ar e the values Of 2 at B an d 0 r e spe c

t ive ly. But by Art . 1 7 7 we have the sam e

'

valu e for 2 whatever

h e the position of T ; thus

P : P ()t”

p.”

v”

) 2ppv 2 g r }x

th erefore O 2pp.v 2 g r }x 2 r }qa .

Thi s h olds th en for every position Of T. Suppose T is at an yp oin t of the g reat circle Of which A is the pole then A: 0 thuswe g e t puy 0

,an d therefore p 0. Similarlyq0

,an d r 0.

1 7 9 . Le t there be an y n umber Of fixed poin ts on the surfaceOf a sphere den ote them by H H

2 ,H

3 ,from an y two poin ts

T an d U on t h e sur face of t h e sphere arcs ar e drawn to the fixed

poin ts : it is required to fin d the sum Of t h e products Of the correspon din g cosin es, that is

cos TH,cos UH

,+ cos TH

,cos UH

2+ cos TE

3cos UH

3+

Le t the n otation be t h e same as in Art . 1 7 4 an d let N, u

’

,v'

be the cosin es Of the arcs wh ich join U with A,B, 0 respectively.

Then by Ar t . 1 66,

cos TH,cos UH

, (Al, pm,v n

,) pi’

m,

v’

n,)

h h’

lf -t vv'

n,

”

(Apt/

4 uh’

)l,m,m

,n,

n,l,.

1 34 ARCS DRAWN TO FIXED POI NTSJ

Similar results hold for cos TH,cos UH

,cos THcos UH

Hen ce, with the n otation Of Art . 1 7 4,the requi red sum is

AXP W'

Q W'

R (sv’

14 010 W M’

)s (W M’

)7‘

Now by properly choosin g the position of the trian g le ABOwe have p , q, an d r each zero as in Art . 1 7 4 an d thus t h e

required sum becomes

M ’P pipi

'

Q W’

R.

1 80. Th e result Obtain ed in Art . 1 7 4 may be con sidered

as a particular case of that just g iven ; n amely the case in wh ich

t h e poin ts T an d U coin cide.

1 81 . A sphere is described about a regular polyh edron from

an y two poin ts on the surface Of the sphere arcs ar e drawn to

t h e solid an g les Of t h e polyhedron it is requir ed to fin d t h e sum

Of t h e products of the correspon din g cosin es.

With the n otation Of Art . 1 7 9 we se e th at t h e sum is

M'

P pp w’

R.

An d here P : ofi—J c 52, by Art . 1 7 7 .

Th us the sum w’

) cos TU.

Th us th e sum of th e p r oducts of th e cosin es is equal to th e

p roduct of th e cosi n e of TU in to a th ird of th e n umbe r of th e solid

an g le s of th e r e gular p olyh e dron .

1 82 . Th e result Obtain ed in Ar t . 1 7 7 may be con sidered as

a particular case of that just g iven ; n amely, the case in wh i ch

t h e poin ts T an d U coin cide.

1 83 . If TU is a quadran t th en cos TU is zero,an d t h e

sum Of the products Of the cosin es in Art . 1 81 is zero. Th e

results p : 0, q: 0, r = 0

,ar e easily seen t o be all special e x

amples of th is particular case.

1 36 MISCELLANEOUS PROPOSITIONS.

1 85 . To fin d th e a n g ular dis tan ce be twe e n th e

”

pole s of th e

i n scribed an d circumscr ibed circles of a trian g le .

Le t P den ote the pole Of the in scribed circle,an d Q t h e pole

Of the circumscribed circle of a trian g le ABO then PAB 5A,

by Ar t . 89, an d QAB z S O, by Ar t . 92 ; hen ce

cosPAQ = cos § (B— O);

an d cos P0 cosPA cos 0A sin PA sin QA cos 5 (B

Now ,by Ar t . 62 (se e the fig ure ofArt .

cosPA cos PE cosAE cos r cos (8 a),

sin PE si n r” A”

sin PAE 8 111 5 45

thus

cosPQ = cos B cos r cos (s —0)cose c § A.

Therefore,by Art . 5 4

cosPQ r: cos B cos r cos (s - a)+ sin B s1n r s1 n § (b+ c)cose c g a,

th erefore00

0

8

0

223r

cot r cos (8 a) tan B sin 5 (b c)cosec 5 a.

2 sin a sin b sin ctan R 2

n

3 2

th erefore058

0

223, r 9

1

,

3 cos (s—a) 2 sin sin %bsin 2°

1

Q

Hen ce 1 (sin a sin b sin c)” l

(cot r tan B)”

(by Ar t .

therefore cos”P0 cos

”B sin

”r cos

”

(B r),

sin”P0 sin

”

(R r) cos”B Sin

”r .

MISCELLANEOUS PROPOSITIONS . 1 37

1 86. To fin d th e an g u lar distan ce be twe e n th e pole of th e

circumscribed circle an d th e p ole of on e of th e e scr ibe d circle s ofa

Le t Q den ote the pole of the circumscribed circle,an d Q the

pole of the escribed circle opposite to the an gle A. Then it may

be sh ewn that QBQ 1 5 7x (C — A), an d

cos Q Q I cosB cos r1cos (s — c)— sin R sin r

lsin % (C—A)se c % B

cosB cos r1cos (3 c) Sin R sin r

1Sin % (c a)cosec § b.

Th erefore

COS

sin coi cot r1cos (8 0)— tan R sin % (c a)cosec b

by reducin g as in t h e precedin g Article,the rig h t-h an d member

t h e last equation becomes

2

1

n(sin b+ sin c — sin a);

h en ce 1 (tan R cot rl)2

, (Art . 94)

th erefore cos2

Q Q , cos? R sin

2rl

cos2

(B

QQ Sin g (R r) cos2B sin g r

l.

1 87 . Th e ar e wh ich p asse s th roug h th e middle poin ts of th e

side s of an y tr ian g le up on a g ive n base wi ll me e t th e bas e p roduce d

a t a fixe d p oin t, th e distan ce of wh ich from th e middle p oin t of th e

base is aquadran t .Le t ABO be an y

.

trian g le, E t h e middle poin t of AO, an d F

t h e middle poin t of AB let the arc wh ich j oin s E an d F when

produced meet B0 produced at Q. Th en

sin BQ sin BF Q sin AQ Sin AFQ.

siuBF sin BQ F’

sin AF sin AQ F’

sin BQ sin AQ Fth ereforesin AQ Sin BQ F

’


sin CQ”

sin AQ F0

sin AQ sin CQ F

therefore sin BQ sin O'

Q th erefore BQ C'

Q 7r.

Hen ce if D be the middle poin t of B0

1 88. If th re e arcs be drawn from th e an g le s of a sp h e ricaltr ian g le th roug h an y p oi n t to me e t th e opp os it e side s

,th e p r oducts

of th e alt er n a te se gme n ts of th e side s ar e equal.

similarly

Le t P be an y poin t, an d le t arcs be drawn from th e an g lesA

,B, C passin g through P an d meetin g t h e Opposite sides at

D,E

,F . Then

Sin BD sin BPD sin CD sin CPD

sin BP Sin BDP sin (JP sm C’DP

sin BD sin BPD sin BPt h e re fm

sin CD sin CPD sin CP

Similar expression s may be foun d for8 1 11 AF

an d h en ce it follows Obviously th at

sin BD Sin CE sin AF

sin CD sin AE sin BF

t h erefore sin BD sin CE sin AF sin CD sin AE sin BF.

HO MISCE LLANEOUS PROPOSITIONS.

Sphere,as A an d B

,can be broug h t in to an y oth er position s, as

A'

an d B'

,by rotation roun d an axi s passin g throug h the cen tre Of

the Sphere an d a certain poin t P. Hen ce it may be in ferred thatan y chan g e Of position in a r ig id body, of which on e poin t is fix ed ,may be effected by rotation roun d some axis throug h the fixedpoin t.

(De Morgan ’

s Difi r e n tial an d I n te g ral Calculus, pag e

1 91 . Le t P den ot e an y poin t with in an y p lan e an g le AOB,

an d from P draw perpen diculars on the straig ht lin es 0A an d

0B then it is eviden t th at these perpen diculars in clude an an g le

wh i ch is the supp leme n t of the an g le AOB. Th e correspon din gfact with r espect to a solid an g le is worthy of n otice . Le t therebe a solid an g le formed by three plan e an g les, meetin g at a poin t

0. From an y poin t P within the solid an g le, draw pe rpe n dicu

lars PL,P31

,PN on t h e three plan es wh ich form t h e solid an g le

then the Spherical trian gle which correspon ds to the three plan es

LPM,MPN

,NPL is the p olar tr ia n g le of the spherical trian g le

wh ich corr espon ds to th e solid an g le at 0. This remark is due

to Professor De Morgan .

1 92 . Suppose thr e e straig ht lin es to meet at a poin t an d forma solid an g le le t a

, ,B,an d 7 den ote t h e an g les con tain ed by these

th r ee straig ht lin es taken in pairs : then it h as been proposed to

call the expression J( 1 cosga cos

’

B 2 cos a cos B cos

the sin e of th e solid an g le . S e e Balt z e r’s D e t e rmi

n an te n,2 n d edition

,pag e 1 7 7 . Adoptin g this defin ition it is easy

t o shew that the sin e of a solid an g le lies between zero an d un ity.

We kn ow that the area Of a plan e tri an g le is half the product

Of two sides in to the sin e of the in cluded an g le : by Ar t . 1 5 6 we

h ave t h e followin g an alog ous proposition ; th e volume of a tetra

h e dron is on e Six th Of the product Of th ree edges in to the Sin e of

t h e solid an g le wh ich they form .

Ag ain ,we kn ow in mechan ics that if th ree forces actin g at a

poin t ar e in equilibrium ,each force is as the sin e of the an g le

between th e direction s of t h e other two : the followin g proposition

is an alog ous ; if four forces actin g at a poin t are in equilibrium

MISCELLANEOUS PROPOSITIONS . 1 41

each force is as t h e sin e Of t h e solid an g le formed by the direction s

Of the other three. Se e Sta tics, Chapter II.

1 93. Le t a sph ere be descr ibed about a r eg ular polyhedron ;let perpen diculars be drawn from the cen tre Of the spher e on the

faces of the polyhedron,an d produced to meet the surface of

the sph ere : then it is Obvious from symmetry that the poin ts of

in tersection must be t h e an gular poin ts Of an other reg ular poly

h e dron .

This may be verified. It will be foun d on examin ation that if

S be t h e n umber Of solid an g les, an d F the n umber of faces Of on e

reg ular polyhedron , then an other regular polyhedron ex ists wh i ch

h as S faces an d F solid an g les. S e e Ar t . 1 5 1 .

1 94. Polyh e dron s .

‘ Th e result in Ar t . 1 5 0 was first Obtain ed

by E uler ; the demon stration which is there g iven is due to

Leg en dre . Th e demon stration shews that the result is true inman y cases in which the polyhedron h as r e —e n tran t solid an g les ;for all that is n ecessary for the demon stration is

,th at it sh all be

possible to take a poin t within the polyhedron as the cen tre Of a

Sphere,SO th at the polyg on s, formed as in Art . 1 5 0

,shall n ot have

an y coin ciden t portion s. Th e result,however

,is g en erally true,

even in cases in which t h e con dition r equired by the demon strat ion of Ar t . 1 5 0 is n ot satisfied. We shall accordin g ly g ivean oth er demon stration , an d shall then deduce some import an tcon sequen ces from t h e result. We beg in with a theorem wh ichis due to Cauchy.

1 95 . Le t th e r e be an y n e twork of r e ctilin e a lfig ur e s, n ot n e ce ssari ly in on e p lan e , bu t n ot formi n g a close d surface ; le t E be th e

n umbe r of edg e s , F th e n umbe r qf fig ur e s, an d S th e n umbe r qf’cor n e r p oin ts th e n F S E 1 .

This t h eorem is obviously true in t h e case of a sin g le plan efig ure ; for then F : 1 , an d S z E . It can be shewn to be g en erally true by in duction . For assume the th eorem to be true fora n etwork of F fig ures ; an d suppose that a rectilin eal fig ure Ofn Sides is added to this n etwork

,so that the n etwork an d the

addition al fig ure h ave m sides coin ciden t,an d therefore m 1


corn er poin ts coin ciden t. An d with respect“to t h e n e w

‘

n etworkwhi ch is thus formed, le t F

'

,F

'

,S den ote the same th in gs as

F,F,S wi th respect to t h e Old n etwork. Th en

E'= E + n —m,

1, S

’= S + n

th erefore F'

+ S' — E .

But F + S ==E + 1 , by hypoth esis ; therefore F'

S'= E ’

+ 1 .

1 96. To demon strate E uler ’s theorem we suppose on e face of

a polyhedron removed,an d we thus Obtain a n etwork Of recti

lin eal figur es t o wh ich Cauchy’s theorem is applicable. Thus

F — l

th erefore F +S = E + 2 .

1 97 . I n an y p olyh e dron th e n umber of face s with an Odd

n um ber of sides is e ve n,an d th e n umbe r qf solid an g le s forme d

with an odd n umbe r of p lan e an g les is eve n .

Le t a,b,c,d,

den ote respectively th e n umbers Of faceswh i ch ar e trian g les, quadrilaterals, pen tagon s, hexag on s,Le t a

, B, y, 8, den ote respectively t h e n umbers Of the solid

an g les wh i ch are formed with th r ee,four

,five

,six

,plan e

an g les.

Th en , each edg e belon gs to two faces,an d termin ates at two

solid an gles ; therefore

From th ese relation s it follows th at a c e

a‘

y e are eve n n umbers .

1 98. W ith t h e n otation Of t h e precedin g Article we h ave

From th ese combin ed with t h e former relation s we Obtain2 E

2 E

Th us 2B can n ot be less than 3F,or less th an 3S .

1 44 LUSCE LLANE OUS PROPOSH I ONS.

It appears that,in a like man n er

,four

,an d on ly four

,n ew

reg ular solids can be formed. TO such solids,t h e faces Of whi ch

in tersect an d cross,E uler ’s theorem does n ot apply.

201 . Le t us return to Art . 1 95 , an d suppose e the n umber

of edg es in the boun din g con tour , an d e’ the n umber of edg es

with in it ; also suppose s the n umber Of corn ers in the boun din gcon tour

,an d s

' the n umber wi th in it. Then

t h erefore I e e’

s s’

F .

But

ther e fore 1 e z s F.

We can n ow demon strate an exten sion Of E uler ’s th eorem,

wh ich h as been g iven by Cauchy.

202 . l e t a p olyh e dron be de comp osed in to an y n umbe r of

polyh edron s at p le asur e ; le t B be th e n umbe r th us forme d, S th e

n umbe r of solid an g le s, F th e n umbe r of face s, E th e n umbe r ofe dg es : th e n S

-i

For suppose all the polyhedron s un ited,by startin g with on e

an d addin g on e at a time. Le t e, j,s be respectively the n um

bers Of edg es, faces, an d solid an g les in the fir st ; let e’

,f'

,s’

be

respectively t h e n umbers of edg es, faces , an d solid an g les in t h e

seco n d which are n ot common to it an d the first ; let f"

,s"

be respectively t h e n umbers of edg es, faces, an d solid an g les in

t h e third which ar e n ot common to it an d the first or secon d ;an d so o n . Th en we h ave t h e followin g results

,n amely, t h e fir st

by Ar t . 1 96, an d the oth ers by Art . 201 ;

s +f = e 2,

e'

+ l ,

s"

+f"

e"

+ 1,

By addition,sin ce s + s

'

+ s S, f+f’

+

e e'

+ e” — E

,we Obtain

MISCELLANEOUS EXAMPLES. 1 45

203 . Th e followin g referen ces will be useful to th ose who

study the theory Of polyhedron s. E uler,Novi Comme n tar ii

Vol. I V. 1 7 5 8 Leg en dre, Geome tr ic ;Poin sot

,Jour n a l d e l

’F cole Polyt e ch n ique , Cahier X ; Cauchy

,

Jour n al de l’E cole Polyte ch n iqu e , Cahi er XVI Poin so t an d Bertran d,

Comp te s l’Académi e de s Sci e n ce s

,Vol . XLVI ; Catalan ,

Th eor eme s e t Probléme s de Geometric F léme n tair e Kirkman,Ph i

losop h ical Tran saction s for 1 85 6 an d subsequen t years ; Listin g ,‘Abh an dlun g e n de r K

’

o’

n ig lich e n G'

o’

t tin g e n , Vol. X .

MISCELLANEOUS EXAMPLES.

1 . Fin d t h e locus Of the vertices of all rig ht-an g le d sph erical

trian g les havin g the same hypoten use ; an d from the equationObtain ed

,prove that the locus is a circle when the radius Of th e

Sphere is in fin ite.

2 . AB is an arc Of a g reat circle on t h e surface of a sph ere,0

it s middle poin t : shew that t h e locus Of t h e poin t P,such that

the an g le APO= the an gle BPO, con sists Of two g reat circles at

r ig ht an g les to on e an other . E xplain this wh en the trian g le

becomes plan e.

3. On a g iven arc Of a Sph ere,Spherical trian g les Of equal

area ar e described : shew that the locus Of the an gular poin tOpposite to the g iven arc is defin ed by the equation

tan- l

t an (a8m 6

tan 9 t an 9+ t an

Sm (u + gS)+ tan

Sin (a

wh ere 2 a is t h e len g th Of the g iven arc,0 the arc Of t h e great

circle drawn from an y poin t P in the locus perpen dicular to the

given arc,cf) the in clin ation Of th e g reat circle on which 9 is

T. s. T. L

1 46 MISCELLANEOUS EXAMPLES.

measured to the g reat circle bisectin g t h e g iven arc at rig ht

an g les, an d ,8 a con stan t.

4. I n an y sph erical trian g le

_

cot A cot a + cot ot b

co t a cot b— cosA cos B°

5 . If 9, gb, ll! den ote the distan ces from t h e an g les A,B

,C

respectively of t h e poin t of in tersect ion of arcs bisectin g t h e

an g les Of the sph erical trian g le ABO,shew that

cos 0sin (b

6 . If A'

,B’

,C’

be t h e poles of the sides BC, CA,AB Of a

Sph erical trian g le ABO, shew that the g reat circles AA’

,BB’

, CC"

meet at a poin t P,such that

cos PA cosB0 cos PB cos 0A cos P0 cosAB.

7 . If 0 be t h e poin t Of in t ersection of arcs AD,BE

,OF

drawn from t h e an g les of a trian g le perpen dicular to the Opposit e

sides an d meetin g th em at D,E

,F respectively, sh ew th at

tan AD tan BE tan CF

tan OD tan OF ’tan OF

ar e respectively equal t o

cos A

cos B cos 0’

8 . If p , q, r be t h e arcs of g reat circles drawn from t h e

an g les Of a tr ian g le pe rpen dicular to the opposite side s, (a,(13, (7 , y

'

) the segmen ts in to which these arcs are divided,

shew that

cos p cosq COS r

cos a cos a’

cos [3 cosB’ cosy cos 7

'

1 48 MI SCELLANEOUS EXAMPLES.

of the an g le in clude d by the two Sides, an d E Z is drawn parallelto the tan g en t t o the circle at Q. Shew that the r emain i n g side

Of the Spherical trian g le is equal to the arc Q PZ .

1 6 . If throug h an y poin t P within a sph erical trian g le ABO

g reat cir cles be drawn from the an g ular poin ts A,B, C

' to meetthe opposite sides at a

,b,c respectively

,prove that

sin Pa cos PA sin PbcosPB sin Pc cosP0

sin Aa sin Bb sin Ge

1 7 . A an d Bar e two places on the E arth’s surface on the

same Side of the equat or , A bein g fur ther from th e equator

than B. If the bearin g Of A from B be more n early due E ast

than it is from an y oth er place in the same latitude as B,fin d

th e bearin g of B from A.

1 8 . From the result g iven in example 1 8 Of Ch apter v. in fer

th e possibility of a r egular dodecahedron .

1 9 . A an d B ar e fixed poin ts on the surface Of a sph ere,an d

P is an y poin t o n the surface. If a an d b are g iven con stan ts,shew that a fixed poin t S can always be foun d

,in AB or AB pro

~

duce d,such that

a cosAP bcosBP 8 cos SP,

wh ere s is a con stan t.

20. A,B,

are fixed poin ts on t h e surface Of a Sph ere ;a,b,

ar e g iven con stan ts . If P be a poin t on t h e surface of

t h e sph ere, such that

a cosAP+ 6 cosBP+ c cos CP+ con stan t,

sh ew th at t h e locus of P is a circle.

XVI . NUME RICAL SOLUTION OF SPHE RICAL

TRIANGLE S.

2 04. We sh all give in this Ch apter examples Of th e n ume

r ical solution Of Sph erical Trian g les .

We sh all first take r igh t-an gled trian g les, an d th en Oblique

e s .

205 . Given a = 37°48

’

1 2 b= 5 9°44

’ C : 90°

TO fin d 0 w e h ave

cos c cos a cos b,

L e os 37 ° 48’ 1 2"

98 97 692 7

L cos 5 9°44

’

1 6”

9 702 3945

L cos 0 + 10 1 96 00087 2

0 : 66°32

’

To fin d A we h ave

cot A=Cot a sin b,

L oot 37 ° 48 ' 1 2 1 01 1 02 65 5

L sin 5 9°44

’

1 6 99 3637 70

L cot A 1 0 20-046642 5

A 41°55

'

1 50 NUMERICAL SOLUTION OF SPHERICAL TRIANGLES .

TO fin d B we h ave

cot B cot 6 sin a,

L cot 5 9°

L sin 37°48

’

1 2

L cot B + 10= 1 95 5 34447

B : 70°1 9

’

1 5

206. Given A 5 5°32

’

45 0 c : 98°1 4

'

2 4"

TO fin d a w e h ave

sin a = sin c sin A,

L sin 9 9 95 4932

L sin 5 5°3 2

'

45 9 9 1 62 32 3

L sin a + 1 0 1 99 1 1 7 2 5 5

a 5 4°41

'

TO fin d B w e h ave

cot B= cos c tan A.

Here cos e is n e g ative ; an d therefore cot B wi ll be n egative,an d B greater th an a rig ht an g le. Th e n umerical value Of cos c

is t h e same as that of cos 8 1 ° 45 ' 36

L cos 81°45

’36 9 1 5 63065

L tan 5 5°3 2

’

45 1 01 636 1 02

L cot ( 1 80°- B)+ 1 0 1 9-3 1 991 67

1 80°—B = 4

"

B 101° 47

'

1 5 2 NUMERICAL SOLUTION OF SPHERICAL TRIANGLES.

TO fin d B we havecos A

sm Bcos a

L sin B = 10+L cos A— L cos a,

10 L cos 1 98 397 454

L cos 42°1 8

'

45”

98 6892 89

L sin B = 9 9 7081 65

B : 69°

or 1 1 0°46

’

Oblique -An g le d Trian g les .

Given a : 7 0°1 4

’ b 49°2 4

’

c

use the formula g iven in Art . 45 ,

sin (3 6)Sin (s c)Sin s sin (s —a)

3 : 7 9°1 2

’

20

s — a = 9 5 8

s — b = 2 9°48 10

s — c 40°2 6

'

10”

L sin 2 9°48

’

1 0 9 69637 04

L sin 10 9 8 1 1 97 68

1 95 08347 2

L sin 20 9 9 92 2 465

L sin 8°5 8

’

9 1 92 7 342

1 9 1 849807

1 95 08347 2

1 9 1 849807

2)~3 2 33665

L t an % A— 1 0 1 6 1 6832

5A : 5 5°2 5

'

3s"

A 1 1 0°5 1

'

NUM E RICAL SOLUTION OF SPHERICAL TRIANGLES. 1 5 3

Similar ly to fin d B,

L Sin 8°

9 1 92 7 342

L si n 40°

98 1 1 97 68

1 90047 1 1 0

L Sin 7 9°1 2

’

20 99 92 2 465

L sin 2 9°48

’

10 96 9637 04

1 9 6886 1 69

1 90047 1 1 0

1 9 6886 1 69

2) 1 8 1 60941

L tan %B — 10=

L tan §B =

§ B = 2 4°2 8

'

2

B = 48°5 6

’

4

to fin d 0,

L sin 8°

91 92 7 342

L sin 2 9°48

’

10”

96 9637 04

1 88 89 1 046

L sin 7 9°1 2

'

20 99 92 2 465

L sin 40°2 6

’

10 9 8 1 1 97 68

1 98 042 2 33

1 88 89 1 046

1 9 8 042 2 33

2 1 084881 3

L tan é—0 — l 0 z 1 -5 42 4406

L t an :} 0 = 9 5 42 4406

40 : 1 9°1 3

'

2 4

0 38°2 6

’

1 54 NUMERICAL SOLUTION OF SPHERICAL TRIANGLES .

209. Given a : 68°20

'

2 5,b= 5 2

°1 8

'

1 5 0 : 1 1 7° 1 2

'

20

By Art . 82,

cos % (a b)

tan % (A—B) iglgicot § at (a b) 8

°1’

5 1 9’

20 {5 0 5 8

L OOS 8°1'

5"

99 95 7 335

L cot 5 8°36

'

1 0 9 7 85 5 690

1 9 7 8 1 302 5

L oos 60° 1 9 ' 20 9 6947 1 20

L t an 5~ (A B) 1 00865 905

9 (A +B) 50°40

’

2 8"

L sin 8°1'

5"

91 445 2 80

L cot 5 8°36

’

10 9 7 85 5 690

1 89 300970

L sin 60°1 9

'

20 9 93893 1 6

L tan flA—B): 89 9 1 1 65 4

4(A B) 5° 35

'

Th erefore A 5 6°1 6

'

1 5 B : 45° 4

'

41

If we proceed to fin d c from the formula

sin c

Sin ce sin 0 is g reater than Sin A we sh all Obtain two values for cboth g reater th an a

,an d we shall n ot kn ow wh ich is the value to

be taken .

1 5 6 NUMERICAL SOLUTION OF SPHERICAL TRIANGLES .

L cos 5 2°1 8

'

1 5 9 7 8637 48

L cos 2”

9 1 91 9060

1 89 7 82 808

L cos 30°36

'

33 99 3483 1 9

L cos ( 1 80°

c) 9 0434489

1 80°—c 83

°39

’

1 7”

c 96°20

’

43

Th us by taki n g on ly the n earest n umber of secon ds in th e

tables the two methods g ive values Of c which differ by if,

h owever,we estimate fraction s of a secon d both methods will

agre e in givin g about 434as the n umber of secon ds.

2 10. Given a : 50°45

'

20 5 69°1 2

°

A 44°2 2

'

10

sin bBy Art . 84

,sin a

L sin 69°1 2

'

40 9 9 7 07 62 6

L sin 44°

9 84465 2 5

1 9 8 1 5 41 5 1

L sin 50°45

’

20 98 88995 6

L sin B = 99 2 641 95

B 5 7°34

’

or 1 2 2°2 5

'

I n th i s case there will be two solution s ; se e Art . 86. We

will calculate 0 an d c by Napier ’s an alog ies,

cos 5 (b a)cos 5 (b+ a)

cos (B +A)cos § (B —A)

tan % (b a).

First take the smaller value'

Of B thus

4(B A) 5 0°5 8

'

(B—A) 6°36°

NUMERICAL SOLUTION OF SPHERICAL TRIANGLES. 1 5 7

L cos 9°1 3

’

40

L cot 5 0°5 8

’

9 9 087 5 36

1 9 9030966

L cos 5 9°5 9

'

9 699 1 887

L tan “20 1 9 2 03907 9

{; C’5 7

°5 8

'

0 : 1 1 5°5 7

’

50 8 .

L eos 5 0° 5 8’ 30 ' 7 9 7 99 1039

L tan 5 9°5 9

’

1 9 2 382 689

2 9 03 7 3 7 2 8

L cos 6°

°7 = 99 97 107 2

L t an ée c :

§ c = 47°39

’

c : 9 5°1 8

'

Next take t h e larg er value of B thus

5 (B A) 83°2 3

’

% (B — A) 39°1'

L cos 9°1 3

’

40 99 943430

L cot 83°2 3

’

39 '3 9 0637 2 97

1 9 05 807 2 7

L cos 5 9°5 9

’

9 699 1887

L tan % 0 : 9 35 88840

5 0 : 1 2°5 2

’

l 5”

'8

0 : 2 5°44

’

3 1 ‘6.

L cos 83°2 3

’

39 8 9 0608369

L t an 5 9°5 9

'

1 9 2 382 689

1 9 2 99 105 8

L oos 39° 1 ’ 2 9 8 : 9 8 903494

L tan c 9 4087 5 64

% c 1 4°2 2

’

c 2 8°45

’

1 5 8 EXAMPLES .

Th e studen t can obtain more examples, wh i ch can be easilyverified, from those here worked out

,by in terchan g in g the g iven

an d required quan tities, or by makin g use of the polar trian g le.

EXAMPLES .

1 . Given b z 1 37°3’

48 A : 1 47°2'

5 4 C = 90°.

Results . 0 : 47 ° 5 7’

a : 1 5 6°1 0

’

34 B : 1 1 3°

2 . Given 0 : 6 1°4'

5 6 a = 40°

3 1'

20 U= 90°.

Re sults . b= 5 0°30

' B : 6 1°50

’

A 47 ° 5 4’

2 1

3 . Given A B C : 90°

Re su lts. a z 20°5 4

’ b : 3 1 ° 43’

3 2 1'

38 5 .

4. Given a = 5 9°2 8

’

A 6 6°7'

20 C :

Resul ts . c = 70°2 3

’

42 b : 48°39

'

1 6 B = 5 2°5 0

’

20

or,

c = 109°36

’

b= 1 3 1°2 0

'

44 B = 1 2 7°9'

40

5 . Given c = a = 1 38°

b= 109°

Re sults . B = 1 20°1 5

'

5 7

6. Given 0 A 1 3 1° B 1 2 0

°

Re su lts . 0 : 109°40

’a = 1 2 7

°1 7

’b 1 1 3° 49

'

3 1

7 . Given a z 7 6° 35’

b 50°10

'

30 c= 40°0'

10

Re sults . A= 1 2 1°36

’

2 0,B = 42° 1 5 ’ 1 3 1 5

'

3 .

8. Give n A= 1 2 9° 5'

2 8 B : 1 42°1 2

’

O'= 1 05

°8'

10

Re su lts. a : 1 35°49

’

20 b : 1 44°37

'

1 5 e z 60°4'

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