TEST BANK FOR Trigonometry 5th Edition by Cynthia Y. Young ISBN:978-1119742623 COMPLETE GUIDE ALL CHAPTERS COVERED 100% VERIFIED A+ GRADE ASSURED!!!!NEW LATEST UPDATE!!!!

1

,CHAPTER 1 cn




Section 1.1 Solutions --------------------------------------------------------------------------------
cn cn cn




1
x 1 x
 
c n c n c n cn c n c n c n

1. Solve for x:
c n cn cn c n cn 2. Solve for x:
c n cn cn c n cn



2 360∘ 4 360∘
360∘  2x, so that x 180∘ .
cn cn c n cn c n cn cn cn 360∘  4x, so that x  90∘ .
c n cn c n cn c n cn cn cn




1 x 2
x
3. Solve for x:   4. Solve for x:  
cn c n c n c n c n c n

c n cn cn c n cn cn c n cn cn c n c n cn



3 360∘ 3 360∘
360∘  3x, so that x  120∘ . (Note
cn cn cn cn c n cn cn cn cn 720∘  2(360∘ )  3x, so that x  240∘ .
cn cn cn cn cn cn cn c n cn cn cn cn



: The angle has a negative measure si
cn cn cn cn cn cn cn (Note: The angle has a negative measu
c n cn cn cn cn c n



nce it is a clockwise rotation.)
cn cn cn cn cn re since it is a clockwise rotation.)
cn cn cn cn cn cn




x 5 7 x
 
c n c n c n cn cnc n c n c n

5. Solve for x:
c n cn cn c n cn 6. Solve for x:
c n cn cn c n cn



6 360∘ 12 360∘
1800∘  5(360∘ )  6x, so that x  300∘ .
cn cn cn cn cn cn cn c n cn cn cn 2520∘  7(360∘ ) 12x, so that x  210∘ .
cn cn cn cn cn cn cn c n cn cn cn




4 x x 5
7. Solve for x:   8. Solve for x:  
cn c n c n c n c n c n c n

c n cn cn c n cn cn c n cn cn c n cn cn



5 360∘ 9 360∘
1440∘  4(360∘ )  5x, so that
cn cn cn cn cn cn cn 1800∘  5(360∘ )  9x, so that
cn cn cn cn cn cn cn




x  288∘ .
cn cn cn x  200∘ .
cn cn cn




(Note: The angle has a negative meas
c n cn cn cn cn cn (Note: The angle has a negative measur
c n cn cn cn cn cn



ure since it is a clockwise rotation.)
cn cn cn cn cn cn e since it is a clockwise rotation.)
cn cn cn cn cn cn




9. 10.
a) complement: 90∘ 18∘  72∘ c n cn c n c n a) complement: 90∘ 39∘  51∘ c n cn cn c n c n




b) supplement: 180∘ 18∘  162∘ c n cn c n c n b) supplement: 180∘  39∘  141∘ c n cn cn c n c n




11. 12.
a) complement: 90∘  42∘  48∘ c n cn cn c n c n a) complement: 90∘ 57∘  33∘ c n cn cn c n c n




b) supplement: 180∘  42∘  138∘ c n cn cn c n c n b) supplement: 180∘  57∘  123∘ c n cn cn c n c n




2

, Section 1.1 cn




13. 14.
a) complement: 90∘ 89∘  1∘ c n cn cn c n c n a) complement: 90∘  75∘  15∘ c n cn cn c n c n




b) supplement: 180∘ 89∘  91∘ c n cn cn c n c n b) supplement: 180∘  75∘  105∘ c n cn cn c n c n




15. Since the angles with measures 4x∘ and 6x∘ are assumed to be complemen
c n cn cn cn cn cn c n c n cn cn cn cn cn




tary, we know that 4x∘  6x∘  90∘. Simplifying this yields
cn cn cn cn cn cn cn cn c n cn cn




10x∘  90∘, cn cn cn c n so that x  9. So, the two angles have measures 36∘and 54∘ .
cn c n cn cn c n cn cn cn cn cn c n cn cn




16. Since the angles with measures 3x∘ and 15x∘ are assumed to be supplemen
c n cn cn cn cn cn c n c n cn cn cn cn cn




tary, we know that 3x∘  15x∘ 180∘. Simplifying this yields
cn cn cn cn cn cn cn cn c n cn cn




18x∘ 180∘, so that cn cn cn cn cn x 10. So, the two angles have measures 30∘ and 150∘ .
cn cn c n cn cn cn cn cn c n cn cn cn




17. Since the angles with measures 8x∘ and
c n cn cn cn cn c n cn c n 4x∘ are assumed to be supplementa cn cn cn cn cn




ry, we know that 8x∘  4x∘ 180∘. Simplifying this yields
cn cn cn cn cn cn cn cn c n cn cn




12x∘ 180∘, cn cn c n so that x 15. So, the two angles have measures 60∘ and 120∘ .
cn cn cn cn c n cn cn cn cn cn c n cn cn cn




18. Since the angles with measures 3x 15∘and 10x 10∘are assumed to be co
c n cn cn cn cn c n cn cn c n cn cn cn cn cn cn




mplementary, we know that 3x 15∘  10x 10∘  90∘. Simplifying this yields cn cn cn cn cn cn cn cn cn cn c n cn cn




13x  25∘  90∘, cn cn cn cn cn so that 13x∘  65∘ and thus, x  5. So, the two angles have meas
cn cn cn cn c n cn c n cn cn c n cn cn cn cn cn




ures 30∘and 60∘ .
c n cn cn




19. Since     180∘, we know th
c n cn cn cn cn cn c n cn c n cn cn 20. Since     180∘, we know th
c n cn cn cn cn cn c n cn c n cn cn




at at
1 17∘ –33∘  180∘ and so,   30∘ . 1 10∘ –45∘  180∘ and so,   25∘ .
– –
cn cn cn cn cn cn cn c n cn cn cn cn cn cn cn cn cn cn cn cn cn cn

cn cn


cn150∘ cn155∘



21. Since     180∘, we know th
c n cn cn cn cn cn c n cn c n cn cn 22. Since     180∘, we know th
c n cn cn cn cn cn c n cn c n cn cn




at at
 4      180∘ and so,   30∘.
cn cn cn cn cn cn cn cn cn cn cn cn cn 3     180∘ and so,   36∘.
cn cn cn cn cn cn cn cn cn cn cn cn cn


–– –– –– ––
cn6cn cn5

Thus,   4 120∘ and     30∘ .
c n cn cn c n cn c n cn c n cn c n cn cn Thus,   3 108∘ and     36∘ .
c n cn cn c n cn c n cn c n cn c n cn cn




3

,