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 mf




Section 1.1 Solutions --------------------------------------------------------------------------------
mf mf mf




1 x 1 x
 
m f m f m f m
f m f m f m f

1. Solve for x:
m f mf mf m f mf 2. Solve for x:
m f mf mf m f mf



2 360∘ 4 360∘
360∘  2x, so that x  180∘ .
mf mf m f mf m f mf mf mf 360∘  4x, so that x  90∘ .
mf mf m f mf m f mf mf mf




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

m f mf mf m f mf mf m f mf mf m f m f mf



3 360∘ 3 360∘
360∘  3x, so that x  120∘ . (Not
mf mf mf mf m f mf mf mf mf 720∘  2(360∘ )  3x, so that x  240∘
mf mf mf mf mf mf mf m f mf mf mf



e: The angle has a negative measure
mf mf mf mf mf mf mf . (Note: The angle has a negative mea
mf m f mf mf mf mf m f



since it is a clockwise rotation.)
mf mf mf mf mf sure since it is a clockwise rotation.)
mf mf mf mf mf mf




x 5 x 7
 
m f m f m f mf mfm f m f m f

5. Solve for x:
m f mf mf m f mf 6. Solve for x:
m f mf mf m f mf



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




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

m f mf mf m f mf mf m f mf mf m f mf mf



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




x  288∘ .
mf mf mf x  200∘ .
mf mf mf




(Note: The angle has a negative mea
m f mf mf mf mf mf (Note: The angle has a negative measu
m f mf mf mf mf mf



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




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




b) supplement: 180∘ 18∘  162∘ m f mf m f m f b) supplement: 180∘  39∘  141∘ m f mf mf m f m f




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




b) supplement: 180∘  42∘  138∘ m f mf mf m f m f b) supplement: 180∘  57∘  123∘ m f mf mf m f m f




2

, Section 1.1 mf




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




b) supplement: 180∘  89∘  91∘ m f mf mf m f m f b) supplement: 180∘  75∘  105∘ m f mf mf m f m f




15. Since the angles with measures 4x∘ and
m f mf mf mf mf mf m f m f 6x∘ are assumed to be compleme mf mf mf mf mf




ntary, we know that 4x∘  6x∘  90∘. Simplifying this yields
mf mf mf mf mf mf mf mf m f mf mf




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




16. Since the angles with measures 3x∘ and
m f mf mf mf mf mf m f m f 15x∘ are assumed to be suppleme mf mf mf mf mf




ntary, we know that 3x∘  15x∘  180∘. Simplifying this yields
mf mf mf mf mf mf mf m
f m f mf mf




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




17. Since the angles with measures
m f mf mf mf mf m f 8x∘ and 4x∘ are assumed to be supplement
mf m f mf mf mf mf mf




ary, we know that 8x∘  4x∘  180∘. Simplifying this yields
mf mf mf mf mf mf mf m
f m f mf mf




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




18. Since the angles with measures 3x 15∘ and
m f mf mf mf mf m f mf m
f m f 10x 10∘ are assumed to be co mf m
f mf mf mf mf




mplementary, we know that 3x 15∘  10x 10∘  90∘. Simplifying this yields
mf mf mf mf mf mf mf mf mf mf m f mf mf




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




asures 30∘and 60∘ . m f mf mf




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




hat hat
1 17∘ –33∘  180∘ and so,   30∘ . 1 10∘ –45∘   180∘ and so,   25∘ .
– –
mf mf mf mf m
f mf mf m f mf mf mf mf mf mf mf m
f mf mf mf mf mf mf

mf mf


mf150∘ mf155∘



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




hat hat
 4          180∘ and so,   30∘.
mf m
f mf mf mf mf mf m
f mf mf mf mf mf 3        180∘ and so,   36∘.
mf m
f mf mf mf mf mf m
f mf mf mf mf mf


–– –– –– ––
mf6m
f  mf5

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




3

,