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 ma




Section 1.1 Solutions --------------------------------------------------------------------------------
ma ma ma




1 x 1 x
 
m a m a m a m
a m a m a m a

1. Solve for x:
m a ma ma m a ma 2. Solve for x:
m a ma ma m a ma



2 360∘ 4 360∘
360∘  2x, so that x  180∘ .
ma ma m a ma m a ma ma ma 360∘  4x, so that x  90∘ .
ma ma m a ma m a ma ma ma




1 x 2 x
3. Solve for x:   4. Solve for x:  
m a m a m a m a m a m a

m a ma ma m a ma ma m a ma ma m a m a ma



3 360∘ 3 360∘
360∘  3x, so that x  120∘ . (No
ma ma ma ma m a ma ma ma ma 720∘  2(360∘ )  3x, so that x  240∘
ma ma ma ma ma ma ma m a ma ma



te: The angle has a negative measure
ma ma ma ma ma ma . (Note: The angle has a negative me
ma ma m a ma ma ma ma m a



since it is a clockwise rotation.)
ma ma ma ma ma ma asure since it is a clockwise rotation.)
ma ma ma ma ma ma




5 x 7 x
 
m a m a m a ma mam a m a m a

5. Solve for x:
m a ma ma m a ma 6. Solve for x:
m a ma ma m a ma



6 360∘ 12 360∘
1800∘  5(360∘ )  6x, so that x  300∘ . 2520∘  7(360∘ )  12x, so that x  210∘
ma ma ma ma ma m a ma m a ma ma ma ma ma ma ma m
a ma ma m a ma ma ma



.
4 x 5 x
7. Solve for x:   8. Solve for x:  
m
a m a m a m a m a m a m a

m a ma ma m a ma ma m a ma ma m a ma ma



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




x  288∘ .
ma ma ma x  200∘ .
ma ma ma




(Note: The angle has a negative mea
m a ma ma ma ma ma (Note: The angle has a negative measu
m a ma ma ma ma ma



sure since it is a clockwise rotation.)
ma ma ma ma ma ma re since it is a clockwise rotation.)
ma ma ma ma ma ma




9. 10.
a) complement: 90∘ 18∘  72∘ m a ma m a m a a) complement: 90∘  39∘  51∘ m a ma ma m a m a




b) supplement: m a 180∘ 18∘  162∘ ma m a m a b) supplement: 180∘  39∘  141∘ m a ma ma m a m a




11. 12.
a) complement: 90∘  42∘  48∘ m a ma ma m a m a a) complement: 90∘  57∘  33∘ m a ma ma m a m a




b) supplement: 180∘  42∘  138∘ m a ma ma m a m a b) supplement: 180∘  57∘  123∘ m a ma ma m a m a




2

, Section 1.1 ma




13. 14.
a) complement: 90∘  89∘  1∘ m a ma ma m a m a a) complement: 90∘  75∘  15∘ m a ma ma m a m a




b) supplement: 180∘  89∘  91∘ m a ma ma m a m a b) supplement: 180∘  75∘  105∘ m a ma ma m a m a




15. Since the angles with measures 4x∘ and
m a ma ma ma ma ma m a m a 6x∘ are assumed to be complemma ma ma ma ma




entary, we know that 4x∘  6x∘  90∘. Simplifying this yields
ma ma ma ma ma ma ma ma m a ma ma




10x∘  90∘ , ma ma m
a m a so that x  9. So, the two angles have measures 36∘and 54∘ .
ma m a ma ma m a ma ma ma ma ma m a ma ma




16. Since the angles with measures 3x∘ and
m a ma ma ma ma ma m a m a 15x∘ are assumed to be supplem ma ma ma ma ma




entary, we know that 3x∘  15x∘  180∘. Simplifying this yields
ma ma ma ma ma ma ma m
a m a ma ma




18x∘  180∘, ma m
a ma so that x  10. So, the two angles have measures 30∘ and 150∘ .
ma m a ma m
a m a ma ma ma ma ma m a ma ma ma




17. Since the angles with measures
m a ma ma ma ma m a 8x∘ and 4x∘ are assumed to be supplemen
ma m a ma ma ma ma ma




tary, we know that 8x∘  4x∘  180∘. Simplifying this yields
ma ma ma ma ma ma ma m
a m a ma ma




12x∘  180∘, ma m
a m a so that x  15. So, the two angles have measures 60∘ and 120∘ .
ma m a ma m
a m a ma ma ma ma ma m a ma ma ma




18. Since the angles with measures
m a 3x 15∘ and 10x 10∘ are assumed to be c
ma ma ma ma m a ma m
a m a ma m
a ma ma ma ma




omplementary, we know that 3x 15∘  10x 10∘  90∘. Simplifying this yields
ma ma ma ma ma ma ma ma ma ma m a ma ma




13x  25∘  90∘, so that 13x∘  65∘ and thus, x  5. So, the two angles have
ma ma ma ma m a ma ma ma ma m a ma m a ma ma m a ma ma ma ma ma




measures 30∘and 60∘ . m a ma ma




19. Since       180∘, we know
m a ma ma ma ma ma m a m
a m a ma ma 20. Since       180∘, we know t
m a ma ma ma ma ma m a m
a m a ma ma




that hat
1 17∘ –33∘   180∘ and so,   30∘ . 1 10∘ –45∘   180∘ and so,   25∘ .
– –
ma ma ma ma m
a ma ma m a ma ma ma ma ma ma ma m
a ma ma ma ma ma ma

ma ma


ma150∘ ma155∘



21. Since       180∘, we know
m a ma ma ma ma ma m a m
a m a ma ma 22. Since       180∘, we know t
m a ma ma ma ma ma m a m
a m a ma ma




that hat
 4         180∘ and so,   30∘.
ma m
a ma ma ma ma ma m
a ma ma ma ma ma 3        180∘ and so,   36∘.
ma m
a ma ma ma ma ma m
a ma ma ma ma ma


–– –– –– ––
ma6ma ma5

Thus,   4  120∘ and     30∘
m a m a ma m a m
a m a ma m a ma m a ma ma Thus,   3  108∘ and     36∘ .
m a ma ma m a ma m a ma m a ma m a ma ma



.

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