.--,,!d .:iirlg trig tunctions ol'an acute

ilt tllat':" -, ,Ectti ansr'cr t in table [urnr ) using

Gite a '"-- '

Section 6.6 The Trigonometry of Right Triangles

Use a calculator to fincl the value of each expression'

rounded to four decimal Places'

24. cos'l2o :"'';'

26. cot 57.3' r, i'

28. csc 39' , .:'t,:,

25. tan 40" rr . ,,r

27. sec 40.9" :.1

29. sin 65' :",'.'''

23 through 30 to answer.

31. sin A : 0.4540 -r',i

33. tan 0 : 0.8391 ':.!:

35. sccB - l.12-10

37. sin 4 -- $.906.3

39. tan a : 0.9896 :': ':

41. sin r-v - 0.32153

74..120 lt '------ -- ll

-'{

iriangies slhown and lvrite answers in tatrle

;.i ri?*t to the nearest 100th of a unit' Yerify

sum to 180'and that the three sides satisfy

y) 6he PYthagorean theorem'

18.

))

0es,

Use a calculator to find the acute angle whose

corresponding ratio is given' Round to the nearest l0th

of a degree. For Exercises 31 through 33' use Exercises

16.32. cosB:0.3090

34. cotA :0.6420

36. cscli : 1.5890

38. lanB:9.6'768

40. cosa:0.7408

42. tana : 3.1336

43.

Select an appropriate function to find the angle

indicated (round to l0ths of a degrec)'

44.

18.7 cm

221 yd

47.

14 in

18 rr

i9.5 cm

45.8 m

CHAPTER 6 An lntroduction to Trigonometric Functions

f)rary a right triangle ABC as

shown, using the informationgiven. Then select :rn appropriateratio to find the side indicated.Round to the nearest t00th.

19. LA : 25" :, :r ,

c : 52 rnntfind side a

Sl" /A:32. :ilria : 1.9 ilifind side ll

53. LA: 62.3'b : E2.5 furlongsfind side c

55. sin 25', cos 65'

56. sin 57". cos 33"

57. tan 5", cot 85o

58. sec 40o, csc 50o

tlased on )our ohseryations in Exerr:ises i.i r^ -

the blank so that the functions given ".- -ijl,]t

Exereises 49 to 54B

Use a calculator to evaluate each pair offunctions andcomment on what you notice.

59. sin 47 . cos

- 60. e ,,s ..-. \in

I

61. cot 69o" tan -: ,' 62. csc tl", sec,il

Complete the following tatrles withouf re{'erringtext or using a calculator. o

6J.

Evaluate the following expressions withoaat a ca

using the cofunction relationship and theforms: sec 75" = \/6 + rt; tan 75" = 2 +

64.

s0.

52.

54.

LB : 55"b: 31 ftlind side c

L_8 - 29.6'r: : 9.-5 ydhnd side a

LB : 12.5"a : 32.8 kmfind side l:

b w*ffiHEru* &qf$YM F*ffitrllLiE-,&S

69. The sine of an angle between two sides of a

triangle: sintl =4ab

If the area lt and two sides a and b of a riangle areknown, the sine of the angle between the two sidesis given by the formula shown. Find the angle 0 forthe triangle shown given "4

: 38.9, and use it tosolve the triangle. {Hint: Apply the same conceptto angle 0 or ry.)

lcos070. Illumination of a surface: E =

The lllumination E of a surface by a light source isa measure of the luminous flux per unit area thatreaches the surface. The vah-re of E fin lumens (1m)

!,,irirt;i.:riri :;r-,:r-"::.i r::iir i-r: lrrLr11; ilt lltt lr.;lrlrrh-,i ...!!r:.i.,1:i 1a-,.-.--.- ..

65. {6 csc 15" t'j . :\.:'i 66. csc2 15o

67. cat2 15" i t ::...'',, 68. ru€cot

per square foot.l isgiven by theformula shown,where d is thedistance from thelight source (infeet), I is theintensity of thelight [in candelas(cd)1, and 6 is theargle the lightsource makes withthe vertical. Forreading a book, an

illumination E ofat least 18 lm/ft? isrecommended.

I (o :

A\17,." '\g,.. \,4N

24

d2Assuming the open book is lying on a hotl€

surface, how far away should a light soulcqplacerl if it has an intensity of 90 cd (aboLtt

and the light flux makes an angle of 65" wll

0 = 30o sin 0 cos 0 t:rn s

cos(90" -. 0) ' tan(90' - 0) csc 0 sec S ::

0=45" ; sin4 cos9 tan$. si

cos(90' - B) : tan(90o - 0) : csc 0 sec (}

book's surface (.i.e.,0 : 25")? ;,i,r:r:r.: l ll

,;^ ^r olevation: In 2001, the tallest building in

:::^:;;. the Petronas Tower I in KualaWult" '' - r:-- f < o frl:i."#;rsia. For a person standing 25'e ft

Iitrlr; ;;;" of tt.," tot".i, the ang1e.of."1":':i:i t:

lop "iirr. tower is 89o' How tall is the Petronas

r,,oole o{ depression: A person standing oear the

::.;; to. Eiffel Tower notices a car wreck some^

Ii*.. f** the tower' If the angle of depressron

il;h; person's eyes to the wreck': :2:' how far

'a1], it"tt ,".rdent from the base of the tower?

*eYt*r4S

" ofeievation: For a person *11dly'99 T:Ji ""ri"i"i the base ol the E'iffel Tower' the

';;;;;;"' to the toP of the tower is 7 I '6" '

,iit i, tt'," EiffelTower?

Exercise 7 i '

*".tcise 72.

es ofl rotation::l.,,fioma point 110 fti.due south of theeastern end of a

|i::::1,,97a**ed eas t/w e s t

??. Height of a climber: A local Outdoors Club has' '-

irt,-r',iL"d to the south rim of a large canyon' when

iir"y tp* a climber attempting to scale the ta11er

noirfl"tn face. Knowing the distance between the

sheer walls of the nortiern and southern faces of

the canyon is approximately 175 yd' they attempt

i"."Lp*" the distance remaining for the climbers

to reach the top of the northern rim' Using a

homemade transit, they sight an angle of

a"pt"t.ion of 55' to the bottom of the north face'

uni ungt", of elevation of 24" and 30" to the

climbers ancl top of the northern rim respeclively'

Section 6.6 The Trigonometry of Right Triangles

(a) How high is

the southernrim of the

canyon?(b) How high is

tire northernrim'? (c) Howmuch fartheruntil the climber

lhe ground. From a

clistance of 500 ft the angle

of elevation lo the Pinnacleof the tower is 74.6'. The

angle of elevation to the

restatlrart from the same

vantage Point is 66'5"'How tal1is the CNNTou,er? How far below the

pinnacle of the tower is the

reslallrant located?

607

$;X" of depression: A person, standlng on the

, ;;";i it,* P"tronut Tower L""*t::::"^:::'::iil ,rO pi"points her residence' If the angle ol

s/Wp, i u;o" fro m the p er^s on' s tI

: t'o,

l "l 1 : :t^,t|T.\l,in* f:u away (in feet and in miles) is the

*1i,f"".* from the base of the tower? See

leachesthetop?. .i, ,- ,,.,.,.,,r,:I,rl'r ;.1::1lr''rli. ':1tl-11

"i.1 !'l :1ir:rirl! ' : ' ": t 't '' 'l r,,'

73. bililili"g'*lafii.t From her elevated observatton

post 300 fi away, a naturalist spots a troop ol

ilrUnont high up in a tree' Using the small transit

attached tollerielescope. she finds the angle of

a.fi"*tion to the bottom of thts tree is 14"' while the

orgt" of elevation to the top of the lree is 25"' The

,n!f" of elevation to the troop of baboons is 2l "'

Usl this information to find (a) the height of the

observation post, (b) the height of the baboons' lree'

1ro i.t iry 1;lchlg:,rk ?pogns a!,!y€,891d,,

Zq. lngl" of elevation:'The tallest free-standing lower

in tie world is the CNN Tor'ver in Toronto' Canada'

Thetowerirrclrrdesarotatingrestatlrantlriglrabove

ri1;,bridge spanning the

t,l.'::.lllinois river, a surveyorrnotcs llre western end

rlies on a bearing of,N 38'35'15'w. To the

jl:,.i1earest inch. how lolg

Yill f. bridge be?

Angles of rotation: Alarge sign spans aneastlwet:r highway.From a point 35 m duewest of the southern base of the sign, a surveyol&nds rhe northern base iies on a bearing ofN 6?" I '1'42" E.To the nearest centimeter' horvwide is the stgn? t-, ; , rr,

,|I

N EO.

i,:i r1,... li't:....' :.1 1!'l'' '' I

iret" or "i.ouiio-h-,

ln Jonuuty 2009' Bul.Dubai

.;;"?fi.i;it captnrecl tlre record as the world's rallesi

l,.,if Ai"g, o..oiaing io tl-re Councii on Tali Buildings

ancl Ur-ian H abittrt ('S o tt rc e : www'ctbuh'or g)'

ftf"^tl.,."d at a poiilt 159 rn trom its base ' the lingle

of eierration to ih" top of the spire is 79' Ftotr a

Xi 35m

/rurKffi- lffiIHffi'Mr&q

t, 1' :1 ':':"' '

768 CHAPTER B Applications of Trigonometry

35. Alternative form for the law of cosines:

h2+c2-s)COSA :

2hcBy solving the law ofcosines for the cosine ofihe angle, the formlrla Acan be written as shown.

36. The perimeter of a trapezoid:P=a+b+/z(csca*cscB)

39. Runway length:Surveyors aremeasuring a large,marshy area outside ofthe city as part of aleasibility study for theconstruction of a newairport. Using a

theodolite and themarkers shown gives

The perimeter of atrapezoid can beforind using thelbrmula shown,where a and brepresent the

with base angies a : 42" ancl p : 76. ,:

Derive-this formuia (solve for cos g), beginningfrom ar : b2 + t;t - Zb, cos A. then us*e this formto begin the solution of the triangle given. :,. .. .,1;.:.:,

37. Distance between cities: The satellite Mercury IImeasLtres its distance from portland and fiomGreen Bay using radio waves as shown. Using anon-board sighting device, the satellite determinesthat LM is 99". How many miles is it fromPortland to Gleen Bay? ;)ir;:rr :r',' ..,

Mercury II

Poltland n7 Green Bay

Distance between cities: Voyager VII measures itsdistance from Los Angeles and from San Franciscousing radio waves as shown. Using an onboardsighting device, the satellite deternines LV is 95".How many kilometers separate Los Angeles andSan Francisco?

the information indicated. ll'the mairr rlrnwaybe at least 1i,000 ft long, and environraental',

4$.

38.

on a proposed tunnel through Harvest &4ountaiorder to find the tunnel's length, the mea

(b) Due to previous tunneiing experience, theestimates a cost of $5000 per foot for boring-

management insists ol a25Vo proflt, what wil,l

37 nr

Voyager VII

610 yd

San Franciscr-r Los Angeles

Iengrhs o[ rhe pal.allel sides. ii is rhr. i:..ilhr ^ctrapezoirl, and a anrJ P l:,5,?l:",1.,:-,';:: ;ij:)rlre pcrimetel oi'Trapeztrid Par.k , ,n ,,,. n.r*l',toot) iI a - 5000 lr. h - 7-500 t'r. .nc l, =-r,i],..

concerns are satisfied, can the airport be consiiucat this site (recall thar I rni = 5280 tri.''

.: l

Ttrnnel length: An engineering flrm decides to bi

shown are {aken. (a) How long will the tunoelii$

through this type ofrock and construcring the i

tunnel according to required speciflcations. lf .,

their minimum bid to the nearest huncJred? ,,.

Section 8.2 The Law of Cosines; the Area of a Triangle 769

i'iri": -'ness executive is going to fly6ip ptanningt l: out';,

,:r^r^^ +^ Cnnaoe cove.:ffi;;ffi from P,ovidenc:]:.:::""t:::'"

$, J*i.orx"i'* t':T:::"*::: "T31 l#il;l; ffi ;;i, *^',,. I:' Tr"::1': :l:::"':l-"*:

::jl ;ffi;;ii;"' what is the measure of angre

l',,,' Iri*" heading-sh.ould she set for this trip?li#;-o;i'""Ji1e.;1.9u1d

she set ror this trip?

45. Geoboard geometry: A rubber land ls

11aced on a

-'' ;;";;oiu-uo*o with ali pegs 1 cm apart) as

shown. Approxtmate the perimeter of the triangle

;;;"; uvir'," rubler baid andthe angle formed at

each vertex. (I{lnr Use a standard triangle to find

/*A andlength AB'),' :,': t'"'tt''l "" l:1r''' jl "'ll' l' i'" l1'1!"

Exereise 45

"d;;:;ffig should theY set for this triP?

60Ge€soo0OGGOG&

OGGOcls&&6

Exercise 46

ssa*soaoo0coso6S&46

oocs0oo--\L

MMannerlY Main

c:423mi

College Cove

NorthI

WesrAEastYSouth

: ti4 mI

Providence

46. Geoboard geometry: A rubber band is piaced on a

geoboard u, 't'otn'

ipproximate the perimeter of

the triangle formed UV if'" ruU!31band and the

angt" folm"a at each vertex' (F/irr: Use a

n'?rlrr,".13"t I'il1"l.:}'l l$ irsti 1'l) of Scouts is Planning aTriP Planning: A trooP

lrif.i iro* Montgomery to Pattonvill"' theY -'

,i-f.t U* the distances shown u.si19,a *11:.Yng

*;Jrtr;". for reference since it is due east of

il,,l,fr;;;;;;rrv' what is the measure,gt iill" ''

10 mi ln Exercises 47 and 48, three rods are atlached via pivot

il;;;;il; rods can tre manipulated to fo'm a triangle'

'f#1fr" three angles of the triangle fbrmed'

3r,,.Aerlal distance: Two planes leave LosAngeles

,, -,International Airport ui th" turnt time' One o'l:it^il t;;;A rr"rii*e 270') with a cruising speed of

4t0;;,';;ing tn for.vn. J'po:: witha sroup that

seeks i,"anluiliiy at the ro919r \4.ount I lll- ] :"

olhe, rareis at heading 225' with a cn"rising.-speed

of 421 rrrph. going to Brisbanc- AustraliiL' with a

lgro.rp ,""kin! udrent.,re in the Great Ou.*utl:..^--

Nautieal distance: Two ships leave Htl-ntrlultr

Harbor at the same time. One tlavels 15 knots

in.r,i.,J *ii., p",. t-,out, at lreading ll0''-1Tl itgoinl i,r r5e Mapoucsu< lslanclt (Croshy' Stillt' anJ

't"I.',; i'ii. ottrer-tlavels i2 knots at heading 200"'ano;s going t; the Samoan lsiands (.Samttct',1^e galuq tu). *owiar apart are the two ships after 10 hr?

47.

:tl:.l" [ppl rr:;;d; airtrr." between the planes after

5 hr o{ 11lght. rr,r:r..r :,rl

49. Peniagon Perimeter: Frnd the

apploximate Perinreter oi zi

regLta. Pentctgot'L lht.rl is

.ii.u-*..it'"d bY a circle rvith

ruJitts i' - l0 cm'

50. Hexagon Perimeter: Find the

perimeter of a regular herogott

ihat is circumsciibed bv a

circle with radir:s r : i5 cm'

Exercise 49

,/\/\/ ttt "tt' \II\l\/\./'-\

Hd

\"1t1.I

i

i

1

i

I

7i754

sin 32o sin i8.5"t. 15rl

11.

9.sin 63' sin C

21.9 18.6

sin C sin 19o

ri6.) +3.1

13. side a: J5 cmLA - 38"

/ D -

A1aLD - +It-. ,.,, l1,l' . tr ..., lilll.,: i;:t- r.. ,. I :,,"1

15. side b : 1016 in.LA: 30'/B:60" ,ti.; =

16.

lia::i1{xw*sH i gl[r-:d]{ilt *L:: *h

CHAPTER B Applications of Trigonometry

e" *,{iiq{: iltri"$ *ff * xii)'*;:i !*{i !-;4,*l;

Fill in each blank with the appropriate rvord or phrase'

1. For the law of sines, if two sides and an angle

opposite one sicle are given, this is referred to tts

the i :,':ir::r'ir.r', CaSe, SinCe the SOlutiOn iS in dOubt

until iurther work.

3. For positive k < l, ihe equation sin 0 = /c has two

soir-ttions, one in Qr-iadrant and the other

in Quadrant

5. In your own words, explain why the AAS case

results in a unique solution while the SSA case

does not. Give supporting diagrams.I :!!ir. ::.,.. i.tit I t::..,

e *HV$EL*plrds Y*Liffi sKitLs

Solve each of the following equations for the unknownpart (if possible). Round sides to the nearest hundredth

and degrees to the nearest tenth.

sin 52" sin 30o8._D IL

li . l:'l !.::

sin B sin 105"10.

- : ---'--3.14 6.28

;'r' : 'i': r':

sin 38o sin B

125 190; ; -.' i,' ;.-:. '

ffi Solve each triangle using the law ofsines. Ifthe law offfi sines cannot be used, state why. Draw and label a

triangle or Iabel the triangle given before you begin. Use

a calculator to verify your results.

14. side b : 385 mLB: 47"LA: la&"

ir.) .,:....(_ ,,.'-,-:,, -, -,-. -.iilil.,'lt. r. ,-, -lll

Carefullv reread the section if needed.

can exceed

4. After a triangle is solved, you sl-roulcl nlr,y

to onsure that the :'r:tr'' ; side is oppoii_- arrglc.

6. Explain why no triangle is possible in eai(a. A:34',8 :J3",C: 52, ..

a: !4"b:22"c: 18', .,.,,,

b. A : 42",8 : 57o,C: 81",,.,.,,a=1".b-9",c )2"

---: 19 in.

89 yd

2. Two inviolatc propettics o{'I trir,lrle 1i111 ,,,.

used to help solve the ambiguou'' ':*t" ul.iflangles must slllr] to ' and (b) no ii"l

19. LALB

side c

21. LBside zr

LC

/i -Ln-

side c :/D-LD*

r;{i,1." r:

20.4i ...i:l

I2.9 mi?:45. 2$._ /<o: r5i2mtij :::r- l; ,"" l:i r:ti -a {:: 103.4o: 42.7 km: 19.6' :t :\ '. :',i '. i:

)",

755Section 8.1 0blique Triangles and the Law of Sines

34.

35.

2'l .5 cm

each question and justify your response usrng a ./38" ,but do not solve'

b

DD

llt.9kty,/,"'

,,:^"

r:t,,on AABC wifh LA: 30o and side c : 20 cm'

iri*r,-, r."grh ror side i ytllt,:*:L1Ll.I^u ,,H;;;i ab) row manv triangles can be formed if

;;;.-; : 8 crn? (c) If side a : l2:m, ho;1manv

:,iio*i"^nv triangles can be formed?

;&i;;;'U;b *i* LA : 60" and side c 1 f V: m'

:{;i *fr^r length for side a will produce a right

l6l3ng1e'? (b) How many triangles can be formed if

36.

37.:...i t.1

:.1

I llilriia" "

: B m? (c) if side a : l0 m' how manY

l.gr", can be iormed? (d) If side a : 15 m' how

many triangtes-.can be formed?1,-jl r ::. lr e, j {1. :

g the law of sines and a scaled drawing' If twtt

exist, solve both comPletelY'

l*iide b : 385 m 28.6'1"

490 m

side a :LB:

side & :t:r.l t.rit:ttrirll:

side c :1-A -

s10e al -'.r _lr: . i.

side b :LD-

side a :

36.5 ydo/12.9 yd

10r,6 in.60"15 in.

Z+.9 t<n-,

32.8 km

lhe taw of, sines to determine if no triangle, one

38. B 6o./-{''-/grr'.s/ _q"'/_4,/+/'

q,/u-,./

,, l)

For Exercises 39 to 44, assume the law of sines is being

applied to solve a triangle' Solve for the unknown

,iei" fif possible), therrdetermine if a second angle

i0'-l ; < 180') exists that also satislies the proportion'

25,8 mi30"17.9 wi

,. il':'i

58 mi59"67 nii

30.

7,)

be formed from the€' or two triangles can I

ms given (diagiams may rtot be to scale), then sin 48"

21

sin 57' _35.6

sinA sin 15'ot' ,*o

: 52

sin 60'4$' 32

sin B

If trvo solution.s exist, solve both completely' Note$!!orvhea* marks the side of undetermined length'

-itt ft ,/

39.

41.sin C t.,laL.

40.2

sin B sin 65'<') Lq

ry:Z _ sin B

121 321

382 cm

1_44.

rve the triangle in Exercise 2 using only the law

;;i;.'' then bv using the Y:f ":t-'.1"', ., ^^$;;y the law of sines' Which method was

;li-ie efficient?f :1. f '. t' " I 29'!"' n * It) 'l ttt. lit*' ttl srtics

6. Begin with a2 : bz + c' * 2b' cosA and write

cos A in terms of a, b, and c (solve for cos A)' Why

must b2 + c2 * az < zbc hold in order for a

solution to exist? r

. t,tt- l',,.'- =l''1 ' ''.ir, ,,:- r,1 >2b,.,.

., . *-. 11":":::":l'l' r!':"::.n"'

1.

fr;' .''l:'n '".''li'f."' ' .

ffi Solve using the law of cosines (if possible)' Latrel each

EH triarrgle appropriately before you tregin' Use a

calculator to verify your results'

24. s 25. s

Section 8.2 The Law of Cosines; the Area of a Triangle

6.7 km

C

767

iiu *t .tt *t the law of cosines can-be used to

solution process for each triangle'

30 km

15 mi

12. ves

CC

i$':f ;i,? !e ,, ,.n,

27. side c : 10\,5 in.sideb: 6tzin.sidea: 15rEin.

28. side a:282ftsideb: 129ftside c : 300 ft i1

29. side a: 32.8kmsideb:24.9kmside c : 1Z.4km

30. s

'- : l:lr, /. lil a' -',..

= {i? 5'. 3 * 1j.4". {' - 85,1 "

,1 ,= 119.3",8 - 41.5". ., - l!.1''

8.

10. iloAC

26.

15 yd

: i:11:iiil:lrr.,lr

tli,ili

j. j.tl

.$l:,i.

ach triangle, verify all three forms of the law of

1. of the following equations fur the unknown

4?,',= 52 + 62 - 2(5)(6)cosB i*4i.'r"

2.92 : 15.22 + 9.82 - 2(15.2)(9.8)cos c c'= 57'i"

if,,,: 92 + 12 - 2(9)(?)cos 52' a - :t.24

$ = 122 + 152 - 2(12)(15)cosA A:4i6'

!,K2' = lB22 + 982 - 2082X9s)cos I r '= E6 e

!:reach triangle using the law of cosines. Use a

to verify your results.

:siaea:75cm 22. side b: 385 mLC:67"

sidea:490m;l.'.' {:6.8'. d = 46'1", r = il9{}'8 rlr

l"C = 38o

iic..:zt-ri*i.LB = 36ldea: l2.9mi A.:13.8,'.f .= 1:6.1'.r''- l$rir

2.9 x tOts rrri D

33.sidea:12{1Ydsideb: l2.9ydside c : 9.2yd A - ile'7" ii -'21 ?'. tj: i6.6''

34. side a : 36.5 AUsideb: l2.9AUside c : 22 AU nor Posriblc

Arklitionll anslveL-r cilil bc 1*Llrril in lhe lllsti.ue$l Alls\Ycr AlrlrlLL \

,"4.

.;)

1iU

o/xlal -at L,'C

A

6.8 AU

49 cm

1114;1 *i

208 cm

For Exercises 8-1L, make a sketch of each compass reading' 10.

11.

8. N70"E 9. N10'W 10. s15"E

12. Match each compass direction with a course'

Direction Course

northeast 135"

southeast 225"

northwest 315'southwest 045'

13. A famous Alfred Hitchcock movie is North

by Northwest. This direction is midway be-

tween north and northwest. What course cor-

responds to this direction?

14. From one corner of a triangular plot of land, a

surveyor determines the directions to the

other two comers to be N32"E and 576"8'

What is the measure of the angle formed by

the edges of the plot of land at the corner

where the surveYor is?

In Exercises 1-4, draw a diagram like those in class Exercises 3-6 to show the

path of a ship proceeding on each given course'

ffi t. o7o' 2. 150" 3' 340' 4' 225'

5. i::;:r:];iliL:;-: Ship A sights ship B on a compass bearing of 080'' Make a sketch

and give the compass bearing of ship A from ship B'

6. li,,:'itil;;;lll ShiP X sights shiP Y

on a bearing of 308". What is the

bearing of X from Y?

7. !y'i',;1;t.1;111 An airplane flies on a

course of i 10" at a sPeed of1200 km/h. How far east of its

starting point is it after 2 tr?

8. A hunter walks east for t h and

then north for lj h. What course

should the hunter take to return [o

his starting point? What assumP-

tions do You make to answer the

question? l#h:$efi?jrffig.PointBisl0kmnorthofpointA,andpointCisl0kmfromBonabearingof

060' from B. Find the bearing and distance ol C from A'

11. S40'W ',.

ffirt

362 Chapter Nine

:i

10.

11.

Point S is 4 km west of Point R' and

point I is 4 km southwest of S' Find

ih" b"aring and distance of R from I'i.r',',: i., 1, ,', Traveling at a speed of

10 knots, a shiP Proceeds south from

its port for I-] h and then changes

.ouir" to I30" for I h' At this time'

how far from Port is the shiP?

lr;r,1 11r;,,',: A sailboat ieaves its dock

and proceeds east for 2 mi' lt then

changes course to 205" until it is due

south of its dock. How far south is

this?

t,

13.

travel to reach shiP A?

Two shiPs, A and B,

leave port at the same time' ShiP A

pro"."d, at 12 knots on a course of

b40', *hil" ship B proceeds at 9 knots

on a course of 115'. After 2 h, shiP A

loses power and radios for help" How

far anO on what course must shiP B

14.

ln Exercises 15-18, sketch each plot of land described and find its area'

,,.1;,,:, 1 ,:,11 11 A ship leaves port and sails northwest for t h and then northeast

for 2 h. If it does no, tf'ung" speed' find what course the ship should take to

return directly to port' efJo fina how long this return will take'

.: :; : :. From an iron post, proceed 500 m northeast to the brook' then

300 m east along the b;k to irtf ota miil' then 200 m S15"E to a post on the

edge of Wiggin', noua, una nnuUy along Wiggin's Road back to the iron post'

l::r;;r",, , From a cement marker' proceed 26A m southwest to the river' then

240 msouth along the river to the bridge' then 280 m N40"8 to a sign on the

edge of Sycamore f-u'", ut'O finally along Sycamore Lane back to the cement

marker.

i:.

i:l

!r

ffi 15.

16.

17. r::irr.,r-.,:,,., From the southeast comer of the ""T"1",?"-o:^::*::,::it;nl".""utr*#;J;t,;;';G'l^l':"ft ."3t:'*H"*'i"i:*il:?;;.;1;ilHi""";#;i#;,";';;:r;"E:';l'o*i::,ol::'*::;:'Ji:,:J,[::iffi""il}! -# "*i i,-i,i"rr""r. Bumham Road, and finallv N30'E along

gumtram Road back to the starting pomt'

lg. l,:u;.lrr.ri::! From rhe intersection of Simpson's.Road *yJlt^:T-kT";0il";;;i' *i;;ri' "'

fi ;;;f ii*p' o,'' Ro ad, then : t'.T

-t:'.,'-1?, :J*3;ffi'#i.J,1iii""r'j:r'ffii;ffi;; ;," is reached, and finauy N68'E

along Mulberry Lane back to the starting pomt'

Triangle TrigonometT 363