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 ap




Section 1.1 Solutions --------------------------------------------------------------------------------
ap ap ap




1
x 1 x
 
a p a p a p ap a p a p a p

1. Solve for x:
a p ap ap a p ap 2. Solve for x:
a p ap ap a p ap



2 360∘ 4 360∘
360∘  2x, so that x 180∘ .
ap ap a p ap a p ap ap ap 360∘  4x, so that x  90∘ .
ap ap a p ap a p ap ap ap




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

a p ap ap a p ap ap a p ap ap a p a p ap



3 360∘ 3 360∘
360∘  3x, so that x  120∘ . (Not
ap ap ap ap a p ap ap ap ap 720∘  2(360∘ )  3x, so that x  240∘ .
ap ap ap ap ap ap ap a p ap ap ap ap



e: The angle has a negative measure s
ap ap ap ap ap ap ap (Note: The angle has a negative measu
a p ap ap ap ap a p



ince it is a clockwise rotation.)
ap ap ap ap ap re since it is a clockwise rotation.)
ap ap ap ap ap ap




5
x 7 x
 
a p a p a p ap apa p a p a p

5. Solve for x:
a p ap ap a p ap 6. Solve for x:
a p ap ap a p ap



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




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

a p ap ap a p ap ap a p ap ap a p ap ap



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




x  288∘ .
ap ap ap x  200∘ .
ap ap ap




(Note: The angle has a negative meas
a p ap ap ap ap ap (Note: The angle has a negative measur
a p ap ap ap ap ap



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




9. 10.
a) complement: 90∘ 18∘  72∘ a p ap a p a p a) complement: 90∘ 39∘  51∘ a p ap ap a p a p




b) supplement: 180∘ 18∘  162∘ a p ap a p a p b) supplement: 180∘  39∘  141∘ a p ap ap a p a p




11. 12.
a) complement: 90∘  42∘  48∘ a p ap ap a p a p a) complement: 90∘ 57∘  33∘ a p ap ap a p a p




b) supplement: 180∘  42∘  138∘ a p ap ap a p a p b) supplement: 180∘  57∘  123∘ a p ap ap a p a p




2

, Section 1.1 ap




13. 14.
a) complement: 90∘ 89∘  1∘ a p ap ap a p a p a) complement: 90∘  75∘  15∘ a p ap ap a p a p




b) supplement: 180∘ 89∘  91∘ a p ap ap a p a p b) supplement: 180∘  75∘  105∘ a p ap ap a p a p




15. Since the angles with measures 4x∘ and 6x∘ are assumed to be complemen
a p ap ap ap ap ap a p a p ap ap ap ap ap




tary, we know that 4x∘  6x∘  90∘. Simplifying this yields
ap ap ap ap ap ap ap ap a p ap ap




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




16. Since the angles with measures 3x∘ and 15x∘ are assumed to be supplemen
a p ap ap ap ap ap a p a p ap ap ap ap ap




tary, we know that 3x∘  15x∘ 180∘. Simplifying this yields
ap ap ap ap ap ap ap ap a p ap ap




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




17. Since the angles with measures 8x∘ and
a p ap ap ap ap a p ap a p 4x∘ are assumed to be supplementa ap ap ap ap ap




ry, we know that 8x∘  4x∘ 180∘. Simplifying this yields
ap ap ap ap ap ap ap ap a p ap ap




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




18. Since the angles with measures 3x 15∘and 10x 10∘are assumed to be co
a p ap ap ap ap a p ap ap a p ap ap ap ap ap ap




mplementary, we know that 3x 15∘  10x 10∘  90∘. Simplifying this yields ap ap ap ap ap ap ap ap ap ap a p ap ap




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




ures 30∘and 60∘ .
a p ap ap




19. Since     180∘, we know th
a p ap ap ap ap ap a p ap a p ap ap 20. Since     180∘, we know th
a p ap ap ap ap ap a p ap a p ap ap




at at
1 17∘ –33∘  180∘ and so,   30∘ . 1 10∘ –45∘  180∘ and so,   25∘ .
– –
ap ap ap ap ap ap ap a p ap ap ap ap ap ap ap ap ap ap ap ap ap ap

ap ap


ap150∘ ap155∘



21. Since     180∘, we know th
a p ap ap ap ap ap a p ap a p ap ap 22. Since     180∘, we know th
a p ap ap ap ap ap a p ap a p ap ap




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


–– –– –– ––
ap6ap ap5

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




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