Solution Manual for Trigonometry 5th Edition Cynthia Y. Young

CHAPTER 1 v




Sectionv1.1vSolutionsv--------------------------------------------------------------------------------
1v v xv 1 xv
v v
v v v
1.v v Solvevforvx:v 2.v Solvevforvx:v
2 360∘ 4 360∘
360∘v v2x,v sovthatv xvv180∘v . 360∘v v4x,v sovthatv xvv90∘v .

1v v xv 2v v xv
3.v Solvevforvx:v v v 4.v Solvevforvx:v v v
3 360∘ 3 360∘
360∘vv3x,vsovthatv xvv120∘v.v(Note: 720∘v v2(360∘v)vv3x,v sovthatv xvv240∘v.v(
v Thevanglevhasvavnegativevmeasurevsin Note:v Thevanglevhasvavnegativev measurev
cevitvisvavclockwisevrotation.) sincevitvisvavclockwisevrotation.)


5v xv 7vv xv
v v
v v v
5.v Solvevforvx:v 6.v Solvevforvx:v
6 360∘ 12 360∘
1800∘v v5(360∘v)vv6x,v sovthatv xvv300∘v . 2520∘v v7(360∘v)vv12x,v sovthatv xvv210∘v .

v4v v xv 5v v xv
7.v Solvevforvx:v v v 8.v Solvevforvx:v v v
5 360∘ 9 360∘
1440∘vv4(360∘v)vv5x,v sovthat 1800∘vv5(360∘v)vv9x,v sovthat
xvv288∘v. xvv200∘v.
(Note:v Thevanglevhasvavnegativevmeas (Note:v Thevanglevhasvavnegativevmeasure
urevsincevitvisvavclockwisevrotation.) vsince vitvisva vclockwise vrotation.)




9. 10.
a) complement:v 90∘v 18∘v v 72∘ a) complement:v 90∘v v39∘v v 51∘
b) supplement:v 180∘v 18∘v v 162∘ b) supplement:v 180∘v v39∘v v 141∘

11. 12.
a) complement:v 90∘v v42∘v v 48∘ a) complement:v 90∘v v57∘v v 33∘
b) supplement:v 180∘v v42∘v v 138∘ b) supplement:v 180∘v v57∘v v 123∘




1

,Chapterv 1


13. 14.
a) complement:v 90∘v v89∘v v 1∘ a) complement:v 90∘v v75∘v v 15∘
b) supplement:v 180∘v v89∘v v 91∘ b) supplement:v 180∘v v75∘v v 105∘

15.v Sincevthevanglesvwithvmeasuresv4x∘v andv 6x∘varevassumedvtovbevcomplementar
y,vwevknowvthatv 4x∘vv6x∘vv90∘.v Simplifyingvthisvyields

10x∘  90∘ , so that x  9.
v v vv v v v v v So,vthevtwovanglesvhavevmeasuresv 36∘andv54∘v .

16.v Sincevthevanglesvwithvmeasuresv3x∘v andv 15x∘varevassumedvtovbevsupplementa
ry,vwevknowvthatv3x∘vv15x∘vv180∘.v Simplifyingvthisvyields

18x∘  180∘, so that x  10.
v v v v v v v v So,vthevtwovanglesvhavevmeasuresv 30∘vandv150∘v.

17.v Sincevthevanglesvwithvmeasuresv 8x∘vandv 4x∘varevassumedvtovbevsupplementary
,vwevknowvthatv8x∘vv4x∘vv180∘.v Simplifyingvthisvyields

12x∘  180∘, so that x  15.
v v v v v v v v So,vthevtwovanglesvhavevmeasuresv 60∘vandv120∘v.

18.v Sincevthevanglesvwithvmeasuresv 3xv15∘vandv 10xv10∘varevassumedvtovbevcomp
lementary,vwevknowvthatv3xv15∘vv10xv10∘vv90∘.v Simplifyingvthisvyields
13x  25∘  90∘, so that 13x∘  65∘ and thus, x  5.
v v v v v v v v v v v v v v v So,vthevtwovanglesvhavevmeasu
resv 30∘andv60∘v.

19.v Sincevvvvvv v180∘,v wevknowvtha 20.v Sincevvvvvv v180∘,v wevknowvthat
t 1 10∘v–45∘vvvv180∘vandvso,v v v25∘v.
– v
1 17∘v–33∘vvvv180∘vandvso,v v v30∘v. v155∘
– v
v150∘



21.v Sincevvvvvv v180∘,v wevknowvtha 22.v Sincevvvvvv v180∘,v wevknowvthat
t 3vvvvv  vvv180∘v andvso,vv v36∘.
 4  vvvvv  vvv180∘v andvso,vv v30∘. –– ––
v5
–– ––
v6v
Thus,v v v3v v108∘v andvv vv v36∘v .
Thus,v v v4v v120∘v andvv vv v30∘v .


2

, Sectionv1.1



23.v vv180∘vv53.3∘vv23.6∘vvv103.1∘ 24.v vv180∘vv105.6∘v13.2∘vvv 61.2∘

25.v Sincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva
v vb v vc .v Usingvthe vgivenvinformation, vthisvbecomesv 4 v v3 v vc v,v whichvsimpli
2 2 2 2 2 2

fiesvtov c2vv25,v sovwevconcludevthatv cvv5. v

26.v Sincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva
v vb v vc .v Usingvthe vgivenvinformation, vthisvbecomes v 3 v v3 v vc v,v which
2 2 2 2 2 2


simplifiesvtov c2v v18,v sovwevconcludevthatv cvv v 18v v3v v 2v .

27.v Sincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva
v vb v vc . v Usingvthe vgivenvinformation, vthisvbecomesv 6 v vb v v10 v, v whichvsimp
2 2 2 2 2 2

lifiesvtov 36vvb2vv100vandvthenvto,vb2vv64,v sovwevconcludevthatv bvv8. v

28.v Sincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva
v vb v vc . v Usingvthe vgivenvinformation, vthisvbecomesv a v v7 v v12 v,v which
2 2 2 2 2 2


simplifiesvtov a2v v95,v sov wevconcludevthatv av  95v.

29.v Sincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva
v vb v vc . v Usingvthe vgivenvinformation, vthisvbecomesv8 v v5 v vc v,v which
2 2 2 2 2 2


simplifiesvtov c2v v89,v sovwevconcludevthatv cv  89v.

30.v Sincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva
v vb v vc . v Usingvthe vgivenvinformation, vthisvbecomesv 6 v v5 v vc v,v which
2 2 2 2 2 2


simplifiesvtov c2v v61,v sovwevconcludevthatv cv  61v.

31.v Sincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva
v vb v vc . v Usingvthe vgivenvinformation, vthisvbecomesv 7 v vb v v11 v,v which
2 2 2 2 2 2


simplifiesvtov b2v v72,v sov wevconcludevthatv bv 72v v6v 2v.

32.vSincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva2
v vb v vc . v Usingvthe vgivenvinformation, vthis vbecomesv a v v5 v v9 v,v which
2 2 2 2 2


simplifiesvtov a2v v56,v sov wevconcludevthatv av  56v v2v 14v.




3

, Chapterv 1


33.v Sincev thisv isv av rightv triangle,v wev knowv fromv thev Pythagoreanv Theoremv thatva
 
2v
2
v vb2v vc2.v Usingvthevgivenvinformation,vthisvbecomesv a2v v v 7v v52v,v which

simplifiesvtov a2v v18,v sovwevconcludevthatv avv v 18v v3v v 2v .

34.vSincevthisvisvavrightvtriangle,vwevknowvfromvthevPythagoreanvTheoremvthatva2
v vb v vc . v Usingvthe vgivenvinformation, vthis vbecomes v5 v vb v v10 v,v which
2 2 2 2 2


simplifiesvtov b2v v75,v sovwevconcludevthatv bv 75v v5v 3v.

35.v Ifv xvv10vin.,v thenvthevhypotenusev 36.v Ifv xvv8vm,v thenvthevhypotenusevofvthis
ofvthisvtrianglevhasvlength trianglevhasvlengthv 8v 2vv11.31vmv.
v

10v 2v v14.14vin.

37.v Letvxvbevthevlengthvofvavlegvinvthevgivenv 45∘vv45∘vv90∘vtriangle.v Ifvthevhypot
enusevofvthisvtrianglevhasvlengthv 2v 2v cm,v then
2vxvv2v 2,vsovthatvxvv2.v
Hence,vthevlengthvofveachvofvthevtwovlegsvisv 2vcmv.

38.vLetvxvbevthevlengthvofvavlegvinvthevgivenv45∘vv45∘vv90∘v triangle.v Ifvthevhypotenuse
v 10v 10v
ofvthisvtrianglevhasvlength 10v ft.,v then 2vxvv v 10,vsovthatvxvv   5.
2 2
Hence,vthevlengthvofveachvofvthevtwovlegsvis 5v ft.

39.v Thevhypotenusevhasvlength 40.vSince 2xvv6mv v xvv 6v 2v v3v 2m,
 
2
2v 4 v v 2v in.vv8vin. eachvlegvhasvlengthv 3v 2v m.

41.v Sincev thev lengthsv ofv thev twov legsv ofv thev givenv30∘v v60∘v v90∘v trianglev arev xv andv3v
x,v thev shorterv legv mustv havev lengthv x.v Hence,v usingv thev givenv information,v we
knowvthatv xvv5vm.v Thus,vthevtwovlegsvhavevlengthsv 5vmvandv 5v 3vv8.66vm,v andvthev
hypotenusevhasvlengthv10vm.

42.v Sincev thev lengthsv ofv thev twov legsv ofv thev givenv 30∘v v60∘v v90∘v trianglev arev xv andv3v
x,v thev shorterv legv mustv havev lengthv x.v Hence,v usingv thev givenv information,v we
knowvthatv xvv9vft.v Thus,vthevtwovlegsvhavevlengthsv 9vft.vandv 9v 3vv15.59vft.,v andvthe
vhypotenuse vhasvlengthv18vft.




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