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!!!!

CHAPTER 1 br




Section 1.1 Solutions --------------------------------------------------------------------------------
br br br




1 x 1 x
 
b r b r b r br b r b r b r

1. Solve for x:
b r br br b r br 2. Solve for x:
b r br br b r br



2 360∘ 4 360∘
360∘  2x, so that x 180∘ .
br br b r br b r br br br 360∘  4x, so that x  90∘ .
b r br b r br b r br br br




1 x 2 x
3. Solve for x:   4. Solve for x:  
b r b r b r b r b r b r

b r br br b r br br b r br br b r b r br



3 360∘ 3 360∘
360∘  3x, so that x  120∘ . (Note
br br br br b r br br br br 720∘  2(360∘ )  3x, so that x  240∘ . (
br br br br br br br b r br br br br



: The angle has a negative measure si
br br br br br br br Note: The angle has a negative measur
b r br br br br b r



nce it is a clockwise rotation.)
br br br br br e since it is a clockwise rotation.)
br br br br br br




5 x 7 x
 
b r b r b r br brb r b r b r

5. Solve for x:
b r br br b r br 6. Solve for x:
b r br br b r br



6 360∘ 12 360∘
1800∘  5(360∘ )  6x, so that x  300∘ .
br br br br br br br b r br br br 2520∘  7(360∘ ) 12x, so that x  210∘ .
br br br br br b r br b r br br br




4 x 5 x
7. Solve for x:   8. Solve for x:  
br b r b r b r b r b r b r

b r br br b r br br b r br br b r br br



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




x  288∘ .
br br br x  200∘ .
br br br




(Note: The angle has a negative meas
b r br br br br br (Note: The angle has a negative measur
b r br br br br br



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




9. 10.
a) complement: 90∘ 18∘  72∘ b r br b r b r a) complement: 90∘ 39∘  51∘ b r br br b r b r




b) supplement: 180∘ 18∘  162∘ b r br b r b r b) supplement: 180∘ 39∘  141∘ b r br br b r b r




11. 12.
a) complement: 90∘  42∘  48∘ b r br br b r b r a) complement: 90∘ 57∘  33∘ b r br br b r b r




b) supplement: 180∘  42∘  138∘ b r br br b r b r b) supplement: 180∘ 57∘  123∘ b r br br b r b r




1

,Chapter 1 br




13. 14.
a) complement: 90∘ 89∘  1∘ b r br br b r b r a) complement: 90∘ 75∘  15∘ b r br br b r b r




b) supplement: 180∘ 89∘  91∘ b r br br b r b r b) supplement: 180∘  75∘  105∘ b r br br b r b r




15. Since the angles with measures 4x∘ and 6x∘ are assumed to be complement
b r br br br br br b r b r br br br br br




ary, we know that 4x∘ 6x∘  90∘. Simplifying this yields
br br br br br br br br b r br br




10x∘  90∘, br br b
r b r so that x  9. So, the two angles have measures 36∘and 54∘ .
br b r br br b r br br br br br b r br br




16. Since the angles with measures 3x∘ and 15x∘ are assumed to be supplement
b r br br br br br b r b r br br br br br




ary, we know that 3x∘  15x∘ 180∘. Simplifying this yields
br br br br br br br br b r br br




18x∘ 180∘, so that br br br br b r x 10. So, the two angles have measures 30∘ and 150∘ .
br br b r br br br br br b r br br br




17. Since the angles with measures 8x∘ and 4x∘ are assumed to be supplementar
b r br br br br b r br b r br br br br br




y, we know that 8x∘  4x∘ 180∘. Simplifying this yields
br br br br br br br br b r br br




12x∘ 180∘, br br b r so that x 15. So, the two angles have measures 60∘ and 120∘ .
br b r br br b r br br br br br b r br br br




18. Since the angles with measures 3x 15∘and 10x 10∘are assumed to be com
b r br br br br b r br b
r b r br b
r br br br br




plementary, we know that 3x 15∘  10x 10∘  90∘. Simplifying this yields
br br br br br br br br br br b r br br




13x 25∘  90∘, br br br br b r so that 13x∘  65∘ and thus, x  5. So, the two angles have measu
br br br br b r br b r br br b r br br br br br




res 30∘and 60∘ .
b r br br




19. Since     180∘, we know th
b r br br br br br b r br b r br br 20. Since     180∘, we know tha
b r br br br br br b r br b r br br




at t
1 17∘ –33∘  180∘ and so,  30∘ . 1 10∘ –45∘  180∘ and so,   25∘ .
– –
br br br br br br br b r br br br br br br br br br br br br br br

br br


br150∘ br155∘



21. Since     180∘, we know th
b r br br br br br b r br b r br br 22. Since     180∘, we know tha
b r br br br br br b r br b r br br




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


–– –– –– ––
br6br br5

Thus,   4 120∘ and     30∘ .
b r b r br b r br b r br b r br b r br br Thus,   3 108∘ and     36∘ .
b r b r br b r br b r br b r br b r br br




2

, Section 1.1br




23.  180∘ 53.3∘  23.6∘  103.1∘
b r br br br br br br br br br 24.  180∘ 105.6∘ 13.2∘  61.2∘
b r br br br br br br br br




25. Since this is a right triangle, we know from the Pythagorean Theorem that a
b r br br br br br br br br br br br br br



2
 b2  c2. Using the given information, this becomes 42 32  c2, which simpl
br br br br b r br br br br br b r br br br br br b r br




ifies to c2  25, so we conclude that c  5.
br b r br br b r br br br b r br br br




26. Since this is a right triangle, we know from the Pythagorean Theorem that
b r br br br br br br br br br br br br




a2  b2  c2. Using the given information, this becomes 32  32  c2, which simp
br br b r br b r br br br br br b r br br b r br br b r br




lifies to c2 18, so we conclude that c 
br b r b r br b r br br br b r br b r b r 18  3 2 . br br b r br




27. Since this is a right triangle, we know from the Pythagorean Theorem that a
b r br br br br br br br br br br br br br



2
 b2  c2. Using the given information, this becomes 62  b2 102, which sim
br br br br b r br br br br br b r br br br br br b r br




plifies to 36  b2 100 and then to, b2  64, so we conclude that b  8 .
br b r br br br br br br br br br br b r br br br b r br br br




28. Since this is a right triangle, we know from the Pythagorean Theorem that
b r b r b r b r b r b r b r br b r b r b r b r b r




a2  b2  c2. Using the given information, this becomes a2  72 122, which
br br b r br b r br br br br br b r br br b r br br b r




simplifies to a2  95, so we conclude that a  95 .
br br b r br br b r br br br b r br br




29. Since this is a right triangle, we know from the Pythagorean Theorem that
b r b r b r b r b r b r b r br b r b r b r b r b r




a2  b2  c2. Using the given information, this becomes 82  52  c2, which
br br b r br b r br br br br br b r br br b r br br b r br




simplifies to c2  89, so we conclude that c  br b r br br b r br br br b r br 89 . br




30. Since this is a right triangle, we know from the Pythagorean Theorem that
b r b r b r b r b r b r b r br b r b r b r b r b r




a2  b2  c2. Using the given information, this becomes 62  52  c2, which
br br b r br b r br br br br br b r br br b r br br b r br




simplifies to c2  61, so we conclude that c  br b r br br b r br br br b r br 61 . br




31. Since this is a right triangle, we know from the Pythagorean Theorem that
b r b r b r b r b r b r b r br b r b r b r b r b r




a2  b2  c2. Using the given information, this becomes 72  b2 112, which
br br b r br b r br br br br br b r br br b r br br b r




simplifies to b2  72, so we conclude that b  72  6 2 .
br br b r br br b r br br br b r br br br b r br




32. Since this is a right triangle, we know from the Pythagorean Theorem that
br br br br br br br br br br br br br




a2  b2  c2. Using the given information, this becomes a2  52  92, which
br br b r br b r br br br br br b r br br b r br br b r br




simplifies to a2  56, so we conclude that a  br b r br br b r br br br b r br 56  2 14 . br br b r br




3

, Chapter 1 br




33. b r Since this is a right triangle, we know from the Pythagorean Theorem that
b r b r b r b r b r b r br b r b r b r b r b r




 7
2b
a2  b2  c2. Using the given information, this becomes a2   52, which simpli
r


br br b r br b r br br br br br b r br br br br br b r br
b r




fies to a2 18, so we conclude that a 
br b r b r br b r br br br b r br b r b r 18  3 2 . br br b r br




34. Since this is a right triangle, we know from the Pythagorean Theorem that
br br br br br br br br br br br br br




a2  b2  c2. Using the given information, this becomes 52  b2 102, which
br br b r br b r br br br br br b r br br b r br br b r




simplifies to b2  75, so we conclude that b  75  5 3 .
br br b r br br b r br br br b r br br br b r br




35. If x 10 in., then the hypotenuse
b r b r br br br b r br br br 36. If x  8 m, then the hypotenuse of th
b r b r br br br b r br br br br




of this triangle has length is triangle has length 8 2 11.31 m .
br br br br

br br br b r b r br br br br



10 2 14.14 in.
b r br br br




37. Let x be the length of a leg in the given 45∘  45∘ 90∘ triangle. If the hypo
b r br br br br br br br br br br b r br br br br br b r br br




tenuse of this triangle has length 2 2 cm, then
br br br br br b r b r b r b r




2 x  2 2, so that x  2.
br br br b r br br br br br br




Hence, the length of each of the two legs is 2 cm .
br br br br br br br br br b r br br




38. Let x be the length of a leg in the given 45∘  45∘ 90∘ triangle. If the hypotenuse
br br br br br br br br br br br br br br br br b r b r br br



10 10
of this triangle has length 10 ft., then 2 x  10, so that x    5.
b r b r br

br br br br br br br br b r b r br br br br br



2 2
Hence, the length of each of the two legs is br br br br br br br br br 5 ft. br




39. The hypotenuse has length
b r br br br
40. Since br 2x  6m  x  br br br br br br
6 2
b r b r  3 2m,
br b r




 
2
2 4 2 in.  8in.
br b r b r br br br br
each leg has length 3 2 m. br br br br b r b r




41. Since the lengths of the two legs of the given30∘  60∘  90∘ triangle are x an
b r br br br br br br br br br br b r br b r br b r br br br




d 3 x, the shorter leg must have length x. Hence, using the given information,
br br b r b r b r b r b r b r br b r b r b r b r b r




we b r




know that x  5 m. Thus, the two legs have lengths 5 m and 5 3  8.66 m, and t
br b r br br br b r br br br br br b r br br b r b r br br br b r br




he hypotenuse has length 10 m.
br br br br br




42. Since the lengths of the two legs of the given 30∘  60∘  90∘ triangle are x a
b r br br br br br br br br br b r br br b r br b r br br br




nd 3 x, the shorter leg must have length x. Hence, using the given informatio
br br b r b r b r b r b r b r br b r b r b r b r b r




n, we b r




know that x  9ft. Thus, the two legs have lengths 9 ft. and 9 3 15.59 ft., and t
br b r br br br b r br br br br br b r br br b r b r br br br b r br




4