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Solution Manual for Trigonometry 5th Edition by Cynthia Y. Young

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Review essential trigonometry concepts with Trigonometry, 5th Edition by Cynthia Y. Young. This solutions resource supports the study of trigonometric functions, identities, equations, graphs, inverse trigonometric functions, and applications of triangles. Use it alongside the textbook and course materials to work through mathematical exercises, understand solution methods, reinforce key formulas, and strengthen your problem-solving skills in trigonometry.

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Solution Manual for Trigonometry, 5th Edition by Cynthia Y. Young
CHAPTER 1
Section 1.1 Solutions --------------------------------------------------------------------------------
1 x 1 x
1. Solve for x:  2. Solve for x: 
2 360∘ 4 360∘
360∘  2x, so that x  180∘ . 360∘  4x, so that x  90∘ .

1 x 2 x
3. Solve for x:   4. Solve for x:  
3 360∘ 3 360∘
360∘  3x, so that x  120∘ . 720∘  2(360∘ )  3x, so that x  240∘ .
(Note: The angle has a negative (Note: The angle has a negative
measure since it is a clockwise measure since it is a clockwise rotation.)
rotation.)
5 x 7 x
5. Solve for x:  6. Solve for x: 
6 360∘ 12 360∘
1800∘  5(360∘ )  6x, so that x  300∘ . 2520∘  7(360∘ )  12x, so that x  210∘ .

4 x 5 x
7. Solve for x:   8. Solve for x:  
5 360∘ 9 360∘
1440∘  4(360∘ )  5x, so that 1800∘  5(360∘ )  9x, so that
x  288∘ . x  200∘ .
(Note: The angle has a negative (Note: The angle has a negative
measure since it is a clockwise measure since it is a clockwise rotation.)
rotation.)

9. 10.
a) complement: 90∘ 18∘  72∘ a) complement: 90∘  39∘  51∘
b) supplement: 180∘ 18∘  162∘ b) supplement: 180∘  39∘  141∘

11. 12.
a) complement: 90∘  42∘  48∘ a) complement: 90∘  57∘  33∘
b) supplement: 180∘  42∘  138∘ b) supplement: 180∘  57∘  123∘



1

Solution manual for trigonometry
Solution
5th edition
manual by
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for
cynthia
trigonometry
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Chapter 1


13. 14.
a) complement: 90∘  89∘  1∘ a) complement: 90∘  75∘  15∘
b) supplement: 180∘  89∘  91∘ b) supplement: 180∘  75∘  105∘

15. Since the angles with measures  4x ∘ and  6x ∘ are assumed to be
complementary, we know that  4x ∘   6x ∘  90∘. Simplifying this yields

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

16. Since the angles with measures 3x ∘ and 15x ∘ are assumed to be
supplementary, we know that 3x ∘  15x ∘  180∘. Simplifying this yields

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

17. Since the angles with measures 8x ∘ and  4x ∘ are assumed to be
supplementary, we know that  8x ∘   4x ∘  180∘. Simplifying this yields

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

18. Since the angles with measures 3x 15 ∘ and 10x 10 ∘ are assumed to be
complementary, we know that 3x 15 ∘  10x 10 ∘  90∘. Simplifying this yields
13x  25 ∘  90∘ , so that 13x ∘  65∘ and thus, x  5. So, the two angles have
measures 30∘ and 60∘ .

19. Since       180∘ , we know 20. Since       180∘ , we know
that that
117∘ 33∘    180∘ and so,   30∘ . 110∘ 45 ∘    180∘ and so,   25∘ .
– – – –
 150∘  155∘



21. Since       180∘ , we know 22. Since       180∘ , we know
that that
 4          180∘ and so,   30∘. 3         180∘ and so,   36∘.
–– –– –– ––
 6   5

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


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Solution manual for trigonometry
Solution
5th edition
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for
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trigonometry
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Section 1.1



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

25. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 42  32  c2 , which
simplifies to c2  25, so we conclude that c  5.

26. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 32  32  c2 , which
simplifies to c2  18, so we conclude that c  18  3 2 .

27. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 62  b2  102 , which
simplifies to 36  b2  100 and then to, b2  64, so we conclude that b  8.

28. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes a2  72  122 , which
simplifies to a2  95, so we conclude that a  95 .

29. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 82  52  c2 , which
simplifies to c2  89, so we conclude that c  89 .

30. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 62  52  c2 , which
simplifies to c2  61, so we conclude that c  61 .

31. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 72  b2  112 , which
simplifies to b2  72, so we conclude that b  72  6 2 .

32. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes a2  52  92 , which
simplifies to a2  56, so we conclude that a  56  2 14 .




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5th edition
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trigonometry
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Chapter 1


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

 7
2
a 2  b 2  c 2 . Using the given information, this becomes a 2   5 2 , which

simplifies to a2  18, so we conclude that a  18  3 2 .

34. Since this is a right triangle, we know from the Pythagorean Theorem that
a2  b2  c2. Using the given information, this becomes 52  b2  102 , which
simplifies to b2  75, so we conclude that b  75  5 3 .

35. If x  10 in., then the hypotenuse 36. If x  8 m, then the hypotenuse of
of this triangle has length
this triangle has length 8 2  11.31 m .
10 2  14.14 in.

37. Let x be the length of a leg in the given 45∘  45∘  90∘ triangle. If the
hypotenuse of this triangle has length 2 2 cm, then 2 x  2 2, so that x  2.
Hence, the length of each of the two legs is 2 cm .

38. Let x be the length of a leg in the given 45∘  45∘  90∘ triangle. If the hypotenuse
10 10
of this triangle has length 10 ft., then 2 x  10, so that x    5.
2 2
Hence, the length of each of the two legs is 5 ft.

39. The hypotenuse has length 40. Since 2x  6m  x  6 2
 3 2m,
 
2
2 4 2 in.  8 in. each leg has length 3 2 m.

41. Since the lengths of the two legs of the given 30∘  60∘  90∘ triangle are x and
3 x, the shorter leg must have length x. Hence, using the given information, we
know that x  5 m. Thus, the two legs have lengths 5 m and 5 3  8.66 m, and
the hypotenuse has length 10 m.

42. Since the lengths of the two legs of the given 30∘  60∘  90∘ triangle are x and
3 x, the shorter leg must have length x. Hence, using the given information, we
know that x  9 ft. Thus, the two legs have lengths 9 ft. and 9 3  15.59 ft., and
the hypotenuse has length 18 ft.



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Solution manual for trigonometry
Solution
5th edition
manual by
Page
for
cynthia
trigonometry
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y young.(exthttps://online2pdf.com/remove-pdf-header-footer)
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