101+ problem solving

8/20/2019 101+ Problem Solving

http://slidepdf.com/reader/full/101-problem-solving 1/24

101+ Solved Problemsin

Trigonometry



I. Angles and Triangles

1. Minutes to Seconds (Conversion of MS!

Convert the following.

Questions Answers

(1) 89° 11' 15" 89.1875°(2) 12° 15' 0" 12.25°( ) ° 0' .5°(!) 71° 0' 0" 71.008 °(5) !2° 2!' 5 " !2.!1!7°( ) .9 !5° ° 57' 52.2"

(7) 1 .12 !5° 1 ° 7' 2!.!2" (8) 21.5° 21° 0' 0" (9) 59.7892° 59° !7' 21.12" (10) 5.18 ° 5° 11' 0.9 "

". egrees and #adians (Conversion of to # and $ice $ersa!#$er%ises

1. Convert e&%h of the following &ngles given in egrees to r& i&ns. ive *our &nswers%orre%t

to 2 e%i+&l ,l&%es.

&) 2- ) 95- %) 217-

2. Convert e&%h of the following &ngles given in r& i&ns to egrees. ive *our &nswers%orre%t

to 2 e%i+&l ,l&%es.

&) r& i&ns ) 2.! r& i&ns %) 1 r& i&n.

. Convert e&%h of the following &ngles given in r& i&ns to egrees. /o not use &

%&l%ul&tor.

&) 15 ) 5

!. Convert the following &ngles given in egrees to r& i&ns. /o not use & %&l%ul&tor &ngive

*our &nswers &s +ulti,les of .



&) 90- ) 72-

Answers

1. &) 0.5 r& i&ns ) 1. r& i&ns %) .79 r& i&ns.

2. &) 171.89- ) 1 7.51- %) 57. 0-

. &) 12- ) -

!. &) 2r& i&ns ) 2 5r& i&ns

II. Trigonometric %unction

1. A surve*or +e&sures & ist&n%e of 12.!5 + to the &se of & uil ing. he &ngle ofelev&tion to the to, of the uil ing is 7 .78 ° . 3h&t is the height of the uil ing 4

2. he &ngle of e,ression fro+ heli%o,ter to & s,ee ing %&r is 5 °. f the heli%o,ter isfl*ing 00+ & ove the groun how f&r is it fro+ the %&r 4

. A tightro,e w&l6er is & out the %ross & ro,e fro+ & o%6 to the ri ge of & o&t th&t is1.5+ & ove the level of the o%6. he ro,e is 15+ long. At wh&t &ngle of in%lin&tion willthe tightro,e w&l6er e %li+ ing4

!. A ,erson is st&n ing h&lfw&* etween two trees th&t & + &,&rt. he &ngles of elev&tionto the to,s of the trees &re !2° &n °. ow +u%h t&ller is one tree over the other4

5. ro+ & ,oint &w&* fro+ the &se of & uil ing the &ngle of elev&tion is 51°. ro+ &,oint !+ %loser to the uil ing the &ngle of elev&tion in%re&ses to 7°. ow t&ll is the

uil ing 4

. 3hile riving through the +ount&ins *ou noti%e & sign th&t shows & hill with & 11gr& e. 3h&t &ngle oes the ro& +&6e with the hori:ont&l4

7. A l& er with its foot on & hori:ont&l fl&t surf&%e rests &g&inst & w&ll. t +&6es &n &ngleof 0° with the hori:ont&l. he foot of the l& er is !1 ft fro+ the &se of the w&ll. inthe height of the ,oint where the l& er tou%hes the w&ll.

8. A +&n on the e%6 of & shi, is 15 ft & ove se& level. e o serves th&t the &ngle ofelev&tion of the to, of & %liff is 70° &n the &ngle of e,ression of its &se &t se& level is50°. in the height of the %liff &n its ist&n%e fro+ the shi,.

9. he &ngle of elev&tion of the to, of & tree is 0 o fro+ & ,oint 28 ft &w&* fro+ the foot ofthe tree. in the height of the tree roun e to the ne&rest feet.



10. ro+ the to, of & s,ire of height 50 ft the &ngles of e,ression of two %&rs on & str&ightro& &t the s&+e level &s th&t of the &se of the s,ire &n on the s&+e si e of it &re 25°&n !0°. C&l%ul&te the the ist&n%e etween the two %&rs.

Answers;

1) 5 ° 2) 72!+ ) 5.7° !) 0. + 5) 10.!+

) . ° 7) 2!ft. 8) 50ft &n 1 ft. 9) 1 ft. 10) !7ft.

III. Inverse Trigonometric %unction

1.A 25 foot t&ll fl&g,ole %&sts & !2 feet sh& ow. 3h&t is the &ngle th&t the sun hits thefl&g,ole4

<olution; irst r&w & ,i%ture. he &ngle th&t the sun hits the fl&g,ole is the &%ute &ngle&t the to, of the tri&ngle . ro+ the ,i%ture we %&n see th&t we nee to use theinverse t&ngent r&tio.

2. #lise is st&n ing on the to, of & 50 foot uil ing &n s,ots her frien =oll* &%ross thestreet. f =oll* is 5 feet &w&* fro+ the &se of the uil ing wh&t is the &ngle of

e,ression fro+ #lise to =oll*4 #lise>s e*e height is !.5 feet.

<olution; ?e%&use of ,&r&llel lines the &ngle of e,ression is e@u&l to the &ngle &t=oll* or . 3e %&n use the inverse t&ngent r&tio.

. o fin the es%&l&tor>s &ngle of elev&tion we nee to use the inverse sine r&tio.

!. he height of & uil ing is 250 ft. 3h&t is the &ngle of elev&tion fro+ & ,oint on thelevel groun 200 ft &w&* fro+ the &se of the uil ing4

<olution;

he lengths of o,,osite &n the & &%ent legs &re 6nown for &ngle B whi%h is the &ngle



of elev&tion.

B t&nD1 (E,,osite FegA &%ent Feg) t&n D1 250200

G 51. H. Ising inverse fun%tions in the%&l%ul&tor.

5. A ,erson st&n s &t the win ow of & uil ing so th&t his e*es &re 12. + & ove thelevel groun in the vi%init* of the uil ing. An o e%t is 58.5 + &w&* fro+ the uil ing on& line ire%tl* ene&th the ,erson. Co+,ute the &ngle of e,ression of the ,erson>s lineof sight to the o e%t on the groun .

<olution; he &ngle of e,ression of the line of sight is the &ngle B th&t the line of sight+&6es with the hori:ont&l &s shown in the figure to the right. <in%e the groun is levelit is ,&r&llel to &n* hori:ont&l line &n so the &ngle th&t the line of sight +&6es with thegroun is e@u&l to B &s well. As & result we h&ve

t&nB 12. 58.5

B t&nD1 (12. 58.5)

B t&nD1 (0.22)

B 12.!0°

. in the si:e of &ngle &°

• <te, 1 he two si es we 6now &re A &%ent ( 750) &n *,otenuse (8 100).

• <te, 2 Ise Cosine un%tion.

• <te, C&l%ul&te Cos $ A &%ent *,otenuse 750 8 100 0.8

• <te, ! in inverse %os of 0.8 ;

%os &° 750 8 100 0.8 inverse %os of 0.8 . °



7. A girl is fl*ing her 6ite with !2° &ngle of elev&tion. f she 6nows the length of the stringof her 6itewhi%h is 00 + how high is the 6ite4 he tri&ngle for+e is shown on the left.

in $ o,,osite h*,otenuse

<in!2° o,,osite 00E,,osite 00sin!2°E,,osite 00(. 7)E,,osite 201 +

8. Chelse& w&l6e u, & ro& th&t h&s & 20 gr& e (she %oul feel itJ) to get to herf&vorite store. At wh&t &ngle oes the ro& %o+e u, fro+ the groun (&t wh&t &ngle isthe ro& in%line fro+ the groun )4

<olution;

Ke+e+ er th&t the gr& e of & ro& %&n e thought of &s &n *ou usu&ll* see it &s& ,er%ent&ge. <o & 20 gr& e is the s&+e &s & gr& e of L for ever* 20 feet thero& goes u, verti%&ll* it goes 100 feet hori:ont&ll*.

I$. Trigonometric Identities

1. Proving of Identities

1.<i+,lif*; sin A M sin A%os 2 A

sin A M sin A%os2 A

sin A(sin 2 A M %os2 A)N &%tor out sin A.O



sin A P 1N<u stitute; sin 2 A M %os2 A 1.O

sin A

2. &%tori:e; %os 2 A D sin2 A %os2 A

%os2 A D sin2 A %os2 A

%os2 A (1 D sin 2 A)N &%tor out %os2 A.O

%os2 A P %os2 AN<u stitute;1 D sin 2 A %os 2 A.O

%os! A

.<i+,lif*; sin ? M %os2? sin?

sin ? M %os2 ?sin ?

sin2 ?M%os2 ?sin ?

1sin ?N<u stitute; sin 2 ? M %os 2 ? 1.O

%s% ?

!.<i+,lif*; se% B D t&n2B se% B

se% B D t&n2Bse% B se%2BDt&n2Bse% B

1se% BN<u stitute; se%2B D t&n2B 1.O

%os B

5. in the v&lue of 1 M se%2B(1Dsin2B).

1 M se%2B(1Dsin2B)

1 M se%2B P %os2BN<u stitute; 1 D sin2B %os2B.O



1 M 1%os2B P %os2B

1 M 1 2

. D (sin A⋅

se% At&n A) .

1 D (sin A ⋅ se% At&n A)

1 D sin A ⋅ (1%os A)t&n AN<u stitute; se% A 1%os A.O

1 D (t&n At&n A)N<u stitute sin A%os A t&n A.O

1 D 1 0

7.<i+,lif*; sin ? R 1 %os ?

sin ? D 1%os ? sin ?%os ? D 1%os ?

t&n ? D se% ?N<u stitute; sin ?%os ? t&n ? &n 1%os? se% ?.O

&. Srove the i entit*

'. Srove the i entit*

10.Srove the i entit* (1 R %os 2(T))(1 M %os2(T)) 2sin 2(T) R sin ! (T)



1 R %os2(T))(1 M %os2(T)) sin 2(T)N1 M %os2(T)O

sin 2(T)N1 M %os2(T) R sin 2(T) M sin2(T)O

sin 2(T)N1 R sin2(T) M sin 2(T) M %os2(T)O

sin 2(T)N1 R sin2(T) M 1O

sin 2(T)N2 R sin2(T)O

2sin 2(T) R sin ! (T)

".Sum and ifference %ormula

1. he &n . is in the @u& r&nt &n is in the .in .

<olution; irst we nee to fin &n . Ising the S*th&gore&n heore+

+issing lengths &re ! &n 5 res,e%tivel*. <o e%&use it is in the

@u& r&nt &n . Uow use the &,,ro,ri&te for+ul&s.

2.Ising the infor+&tion fro+ 1 fin .

<olution; ro+ the %osine &n sine of &n we 6now th&t &n .



.<i+,lif* .

<olution; #$,&n this using the ifferen%e for+ul& &n then si+,lif*.

!.3h&t is the e$&%t v&lue of sin(105H)4

3e %&n use & su+ &ngle for+ul& noti%ing th&t 105H !5H M 0H.

3e h&ve sin(105H) sin(!5H M 0H) sin(!5H )%os( 0H) M %os(!5H )sin( 0H).

3e 6now the e$&%t v&lues of trig fun%tions for 0H &n !5H.

herefore sin(!5H )%os( 0H) M %os(!5H )sin( 0H) .

his is the e$&%t v&lue e%&use we &re using the r& i%&ls to e$,ress e$&%t s@u&re roots. A e%i+&l &,,ro$i+&tion is 0.9 59.

5.3h&t is the e$&%t v&lue of t&n(15H)4

3e %&n use & ifferen%e &ngle for+ul& noti%ing th&t 15H !5H D 0H.

t&n(15H) t&n(!5H D 0H)

his e$,ression whi%h re,resents the e$&%t v&lue of t&n(15H) %&n e rewritten &sfollows so th&t there is no r& i%&l in the eno+in&tor.

. in the e$&%t v&lue of sin(15 o)

<olution

15 o is not & s,e%i&l &ngle. owever 15 !5 D 0 &n oth !5 &n 0 &re s,e%i&l &ngles.en%e

http://def%28%27/Glossary/glossaryterm.aspx?word=Angle%27,%20500,%20500);

http://www.algebralab.org/lessons/lesson.aspx?file=Trigonometry_TrigSpecialTriangles.xml

http://def%28%27/Glossary/glossaryterm.aspx?word=Square%27,%20500,%20500);



http://def%28%27/Glossary/glossaryterm.aspx?word=Radical%20%27,%20500,%20500);

http://www.algebralab.org/lessons/lesson.aspx?file=Trigonometry_TrigSpecialTriangles.xml



http://def%28%27/Glossary/glossaryterm.aspx?word=Radical%20%27,%20500,%20500);




sin(15 o) sin (!5 o D 0o)

• 3e now use the ifferen%e for+ul& for sine.

sin(!5o

)V%os( 0o

) D %os(!5o

)Vsin( 0o

)• <u stitute the v&lues of sine sn %osine of !5 o &n 0 o in the & ove to o t&in.

sin(15 o) Ns@rt(2) 2ONs@rt( ) 2O D Ns@rt(2) 2ON1 2O

• Co++on eno+in&tor &n f&%toring.

sin(15 o) s@rt(2)Ns@rt( ) D 1O !

7.<i+,lif* %os($ D ,i 2)

<olution

• Ise the ifferen%e for+ul& for %osine to e$,&n the given e$,ression

%os($ D ,i 2) %os $ V %os ,i 2 M sin $ V sin ,i 2

• %os ,i 2 0 &n sin ,i 2 1 hen%e.

%os($ D ,i 2) sin $

8. iven sin $ 1 5 &n sin * D2 &ngle $ is in @u& r&nt &n &ngle * is in@u& r&nt fin the e$&%t v&lue of sin($ M *).

<olution

#$,&n sin($ M *) using the su+ for+ul& of the sine.

sin($ M *) sin $ V %os * M %os $ V sin *

• 3e 6now sin $ ut not %os $ we use the i entit* sin 2$ M %os2$ 1 to fin %os $.

%os $ (M or D) <QK (1 D sin 2$)



• <in%e $ is in @u& r&nt %os $ is neg&tive.

%os $ D <QK (1 D (1 5)2)

•

3e 6now sin * ut not %os * we use the s&+e i entit* &s & ove sin2

* M %os2

* 1to fin %os *.

%os * (M or D) <QK (1 D sin 2*)

• <in%e * is in @u& r&nt %os * is neg&tive.

%os * D <QK (1 D (D2 )2)

D <QK (1 D ! 9)

(D1 )<QK (5)

• 3e now su titute sin $ %os $ sin * &n %os * * their v&lues in the for+ul&& ove.

sin($ M *) sin $ V %os * M %os $ V sin *

N1 5OVND(1 )<QK (5)O M ND(1 5)<QK (2!)OND2 O

ND<QK (5) M <QK (2!)O 15

9. in the e$&%t v&lue of %os 15 o

<olution;• hin6 of two &ngles th&t *ou 6now the v&lue of th&t either

& or su tr&%t &n give *ou 15. here &re +&n*. Ene th&t%o+es to +in is !5 &n 0.

let !5 &n 0Ise the & ove for+ul& for the ifferen%e of the %osine%os 15 %os(!5 D 0) %os !5 %os 0 M sin !5 sin 0



2. #$,ress &s & ,ro u%t.

Answer. 3e h&ve

Uote th&t we use .

. Xerif* the for+ul&

Answer. 3e h&ve

&n

en%e



whi%h %le&rl* i+,lies

. in the re&l nu+ er $ su%h th&t &n

Answer. =&n* w&*s +&* e use to t&%6le this ,ro le+. Fet us use the & ove for+ul&s.3e h&ve

en%e

<in%e the e@u&tion gives &n the

e@u&tion gives . herefore the solutions to the e@u&tion

&re

!. Xerif* the i entit*

Answer. 3e h&ve



Ising the & ove for+ul&s we get

en%e

whi%h i+,lies

<in%e we get

5. Xerif* th&t sin T %os Y

<t&rt * & ing the su+ &n ifferen%e i entities for the sine.



he other three ,ro u%t ‐ su+ i entities %&n e verifie * & ing or su tr&%ting othersu+ &n ifferen%e i entities.

. 3rite %os $ %os 2 $ &s & su+.

Altern&te for+s of the ,ro u%t ‐ su+ i entities &re the su+ ‐ ,ro u%t i entities.

hese i entities &re v&li for egree or r& i&n +e&sure whenever oth si es of the

i entit* &re efine .

7.



<olve for T * & ing the following two e@u&tions &n then ivi ing * 2. <olve for Y *su tr&%ting the two e@u&tions &n then ivi ing * 2.

8. 3rite the ifferen%e %os 8T Z %os 2 T &s & ,ro u%t.

9. in the e$&%t v&lue of sin 75° M sin 15°.

10. #$,ress the ,ro u%t %os( $)sin(2$) &s & su+ of trigono+etri% fun%tions.



$. Solution of Angles

1. #ig t Triangle

1. A 1 foot l& er is le&ning &g&inst & house. t tou%hes the otto+ of & win ow th&t is12 feet in%hes & ove the groun . 3h&t is the +e&sure of the &ngle th&t the l& erfor+s with the groun 4

Fet $ e@u&l the +e&sure of the &ngle thel& er for+s with the groun . A ,i%ture of the ,ro le+ is r&wn to the right.

3e h&ve the si e o,,osite to the &ngle in@uestion &s well &s the h*,otenuse. %&nwrite the un6nown in ter+s of the 6nownusing the efinition of sine;

. irst nee to get ever* thing in ter+s of in%hes; 12 ft 1!! in. so 12 ft

in 150 in. &n 192 in so . Uoti%e the units %&n%el out.[ou shoul &lw&*s get & unitless nu+ er when *ou h&ve & trigono+etri% r&tio. Ising theinverse sin on & %&l%ul&tor get the +e&sure of the &ngle is e@u&l to .

2.

&. woul use the %onverse of the S*th&gore&nheore+ to solve this ,ro le+.

. he %onverse of the S*th&gore&n heore+ tells+e th&t if then the ,ole woule &t & right &ngle with the groun when the string

w&s 17 ft.

<o %o+,ute



&n sin%e the ,ole woul e &t & right &ngle with the grounwhen the string is 17 ft.

. \&il& is fl*ing & 6ite whose string is +&6ing & &ngle with the groun . he 6itestring is 5 +eters long. ow f&r is the 6ite & ove the groun 4

After re& ing the ,ro le+ woul r&w the following ,i%ture;

where h is the height wh&t w&nt to fin . 6now &n &ngle so 6now neeto use & trigono+etri% r&tio to solve this ,ro le+. &+ loo6ing for the si eo,,osite the given &ngle &n 6now the h*,otenuse. Foo6ing &t +*

efinitions see shoul use the sine r&tio to write the un6nown in ter+s of 6nowns.

<o h&ve

or .

<o using +* %&l%ul&tor to %o+,ute sin fin the 6ite is &,,ro$i+&tel* 1 +eters& ove the groun .

!. he ?roo6's &re inst&lling & wi eDs%reen television with & 0Din%h i&gon&l. heirentert&in+ent %enter is !8 in%hes wi e * in%hes high will the television fit in their%urrent entert&in+ent %enter4

After re& ing the ,ro le+ r&w the following;

w&nt to fin to see if the television with & 0Din%h i&gon&l will fit into & re%t&ngle th&t is !8Din%hes * in%hes. %&n use the S*th&gore&n

heore+ to ,ut +* 6nowns in ter+s of +*un6nown;

or



so the television will fit e$&%tl*.

5. A surve*or is 100 +eters fro+ the &se of & &+. he &ngle of elev&tion to the to, ofthe &+ +e&sures . he surve*or's e*eDlevel is 1.7 +eters & ove the groun . inthe height of the &+ to the ne&rest hun re th of & +eter.

After re& ing the ,ro le+ r&w;

<o the to, of the &+ will eist&n%e M 1.7 +. h&ve &n

&ngle &n the si e & &%ent tothe &ngle &n &+ loo6ing for the si e o,,osite the &ngle.

<in%e will use thistrigono+etri% r&tio to solve for .

get;

&+ not @uite one +ust & the ist&n%e the tri&ngle is & ove the groun to get theheight &+ loo6ing for. he height of the &+ is !8.77 + M 1.7 + 50.50 +.

. iven &n &%ute &ngle &n one si e. <olve the right tri&ngle A?C if &ngle A is° &n si e % is 10 %+.

<olution. <in%e &ngle A is ° then &ngle ? is 90° Z ° 5!°.

o fin &n un6nown si e s&* & ,ro%ee &s follows;

1. =&6e the un6nown si e the nu+er&tor of & fr&%tion &n +&6e the 6nown si e theeno+in&tor.

In6nown \nown

2. U&+e th&t fun%tion of the &ngle.



In6nown \nown &

10 sin °

. Ise the trigono+etri% & le to ev&lu&te th&t fun%tion.

In6nown \nown

&10 sin ° .588

!. <olve for the un6nown si e.

& 10 ] .588 %+ 5.88 %+

(Fesson ! of Arith+eti%. )

7. <olve the tri&ngle for si e .

o see the &nswer ,&ss *our +ouse over the %olore &re&.o %over the &nswer &g&in %li%6 "Kefresh" ("Kelo& ").o %onsult the & le %li%6here .

In6nown \nown 10 %os ° .809

10 ] .809 8.09 %+

8. o +e&sure the wi th of & river. wo trees st&n o,,osite one &nother &t ,oints A &n ? on o,,osite &n6s of & river.

/ist&n%e AC &long one &n6 is ,er,en i%ul&r to ?A &n is +e&sure to e 100 feet. Angle AC? is +e&sure to e 79°. ow f&r &,&rt &re the treesL th&t is wh&t is thewi th w of the river4

In6nown\nown

w100 t&n 79°.

w 100 t&n 79°

http://www.themathpage.com/atrig/trigonometric-table.htm

http://www.themathpage.com/arith/multiply-by-powers-of-10.htm#multiply



http://www.themathpage.com/arith/multiply-by-powers-of-10.htm#multiply




100 ] 5.1!5 51!.5 ft

fro+ the & le.

( o +e&sure the height of & fl&g,ole &n for the +e&ning of the&ngle of

elev&tion see the #$&+,le in o,i% .)

9. in the ist&n%e of & o&t fro+ & lighthouse if the lighthouse is 100 +eters t&ll&n the &ngle of e,ression is °.

<olution. he &ngle of e,ression is the &ngle elow str&ight &he& DD hori:ont&l DDth&t &n o erver +ust loo6 in or er to see so+ething elow the o server. hus inor er to see the o&t the lighthouse 6ee,er +ust loo6 own °.

Uow the tri&ngle for+e * the lighthouse &n the ist&n%e of the o&t fro+the lighthouse is rightD&ngle . An sin%e the &ngle of e,ression is ° then the&ltern&te &ngle is &lso °. ( #u%li . 29.)

f is the ist&n%e of & o&t fro+ the lighthouse then 100 %ot ° 9.51! fro+ the & le.

herefore 951.! +eters.

10. iven two si es of & right tri&ngle. <olve the right tri&ngle A?C given th&tsi e % 25 %+ &n si e 2! %+.

<olution. o fin the re+&ining si e & use the S*th&gore&n theore+ ;

&2 M 2! 2 25 2

&2 25 Z 57 !9

& 7.Ue$t to fin &ngle A we h&ve

%os A 2! 9 on +ulti,l*ing e&%h ter+ * !.


http://www.themathpage.com/atrig/trigonometry-of-right-triangles.htm#example

http://www.themathpage.com/abookI/propI-29-30.htm#prop29


http://www.themathpage.com/Alg/pythagorean-distance.htm#hhh


http://www.themathpage.com/atrig/trigonometry-of-right-triangles.htm#example

http://www.themathpage.com/abookI/propI-29-30.htm#prop29


http://www.themathpage.com/Alg/pythagorean-distance.htm#hhh



25 100

.9

(<ee <6ill in Arith+eti%; r&%tions into e%i+&ls .)3e +ust now ins,e%t the & le to fin the &ngle whose %osine is %losest to .9

or sin%e this is & three ,l&%e & le .9 0.3e fin

%os 1 ° .9 1herefore

Angle A 1 °.

in&ll* Angle ? 90° Z 1 ° 7!°.

3e h&ve solve the tri&ngle.

". ,bli-ue Triangle

http://www.themathpage.com/arith/fractions-into-decimals-1.htm


http://www.themathpage.com/arith/fractions-into-decimals-1.htm


101+ problem solving

Documents