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!!!!

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, CHAPTER 1 dt




Section 1.1 Solutions --------------------------------------------------------------------------------
dt dt dt




1 x 1 x
 
d t d t d t dt d t d t d t

1. Solve for x:
d t dt dt d t dt 2. Solve for x:
d t dt dt d t dt



2 360∘ 4 360∘
360∘  2x, so that x 180∘ .
dt dt d t dt d t dt dt dt 360∘  4x, so that x  90∘ .
d t dt d t dt d t dt dt dt




1 x 2 x
3. Solve for x:   4. Solve for x:  
d t d t d t d t d t d t

d t dt dt d t dt dt d t dt dt d t d t dt



3 360∘ 3 360∘
360∘  3x, so that x  120∘ . (Note:
dt dt dt dt d t dt dt dt dt 720∘  2(360∘ )  3x, so that x  240∘ . (
dt dt dt dt dt dt dt d t dt dt dt dt



The angle has a negative measure sin
dt dt dt dt dt dt dt Note: The angle has a negative measur
d t dt dt dt dt d t



ce it is a clockwise rotation.)
dt dt dt dt dt e since it is a clockwise rotation.)
dt dt dt dt dt dt




5 x 7 x
 
d t d t d t dt dtd t d t d t

5. Solve for x:
d t dt dt d t dt 6. Solve for x:
d t dt dt d t dt



6 360∘ 12 360∘
1800∘  5(360∘ )  6x, so that x  300∘ .
dt dt dt dt dt dt dt d t dt dt dt 2520∘  7(360∘ ) 12x, so that x  210∘ .
dt dt dt dt dt dt dt d t dt dt dt




4 x 5 x
7. Solve for x:   8. Solve for x:  
dt d t d t d t d t d t d t

d t dt dt d t dt dt d t dt dt d t dt dt



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




x  288∘ .
dt dt dt x  200∘ .
dt dt dt




(Note: The angle has a negative meas
d t dt dt dt dt dt (Note: The angle has a negative measur
d t dt dt dt dt dt



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




9. 10.
a) complement: 90∘ 18∘  72∘ d t dt d t d t a) complement: 90∘ 39∘  51∘ d t dt dt d t d t




b) supplement: 180∘ 18∘  162∘ d t dt d t d t b) supplement: 180∘  39∘  141∘ d t dt dt d t d t




11. 12.
a) complement: 90∘  42∘  48∘ d t dt dt d t d t a) complement: 90∘ 57∘  33∘ d t dt dt d t d t




b) supplement: 180∘  42∘  138∘ d t dt dt d t d t b) supplement: 180∘ 57∘  123∘ d t dt dt d t d t




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, Section 1.1 dt




13. 14.
a) complement: 90∘ 89∘  1∘ d t dt dt d t d t a) complement: 90∘ 75∘  15∘ d t dt dt d t d t




b) supplement: 180∘ 89∘  91∘ d t dt dt d t d t b) supplement: 180∘  75∘  105∘ d t dt dt d t d t




15. Since the angles with measures 4x∘ and 6x∘ are assumed to be complement
d t dt dt dt dt dt d t d t dt dt dt dt dt




ary, we know that 4x∘  6x∘  90∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt




10x∘  90∘, dt dt d
t d t so that x  9. So, the two angles have measures 36∘and 54∘ .
dt d t dt dt d t dt dt dt dt dt d t dt dt




16. Since the angles with measures 3x∘ and 15x∘ are assumed to be supplement
d t dt dt dt dt dt d t d t dt dt dt dt dt




ary, we know that 3x∘  15x∘ 180∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt




18x∘ 180∘, so that dt dt dt dt d t x 10. So, the two angles have measures 30∘ and 150∘ .
dt dt d t dt dt dt dt dt d t dt dt dt




17. Since the angles with measures 8x∘ and 4x∘ are assumed to be supplementar
d t dt dt dt dt d t dt d t dt dt dt dt dt




y, we know that 8x∘  4x∘ 180∘. Simplifying this yields
dt dt dt dt dt dt dt dt d t dt dt




12x∘ 180∘, dt dt d t so that x 15. So, the two angles have measures 60∘ and 120∘ .
dt d t dt dt d t dt dt dt dt dt d t dt dt dt




18. Since the angles with measures 3x 15∘and 10x 10∘are assumed to be com
d t dt dt dt dt d t dt d
t d t dt d
t dt dt dt dt




plementary, we know that 3x 15∘  10x 10∘  90∘. Simplifying this yields dt dt dt dt dt dt dt dt dt dt d t dt dt




13x 25∘  90∘, dt dt dt dt d t so that 13x∘  65∘ and thus, x  5. So, the two angles have measur
dt dt dt dt d t dt d t dt dt d t dt dt dt dt dt




es 30∘and 60∘ .
d t dt dt




19. Since     180∘, we know tha
d t dt dt dt dt dt d t dt d t dt dt 20. Since     180∘, we know that
d t dt dt dt dt dt d t dt d t dt dt




t 1 10∘ –45∘  180∘ and so,   25∘ .
–
dt dt dt dt dt dt dt dt dt dt dt




1 17∘ –33∘  180∘ and so,  30∘ .
dt


dt155∘
–
dt dt dt dt dt dt dt d t dt dt dt

dt


dt150∘



21. Since     180∘, we know th
d t dt dt dt dt dt d t dt d t dt dt 22. Since     180∘, we know that
d t dt dt dt dt dt d t dt d t dt dt




at 3     180∘ and so,   36∘.
dt dt dt dt dt dt dt dt dt dt dt dt dt




 4      180∘ and so,   30∘. –– ––
dt5
dt dt dt dt dt dt dt dt dt dt dt dt dt


–– ––
dt6dt
Thus,   3 108∘ and     36∘ .
d t dt dt d t dt d t dt d t dt d t dt dt




Thus,   4 d t d t dt d t 120∘ dt d t and    dt d t dt d t  30∘ dt dt .


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