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Complete Solutions Manual — Trigonometry, 5th Edition — Cynthia Y. Young, 2021 — ISBN 9781119742623 — Latest Update 2025/2026 (All Chapters Covered 1–8)

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This verified Solutions Manual for Trigonometry (5th Edition) by Cynthia Y. Young (published 2021, ISBN 9781119742623) offers comprehensive, step-by-step solutions for all exercises and problems presented in the textbook. Tailored for students and instructors in college-level trigonometry courses, the manual supports the development of foundational skills and advanced problem-solving strategies across the full scope of trigonometric topics. The official chapter structure begins with Chapter 1: Right Triangle Trigonometry, followed by Chapter 2: Trigonometric Functions, Chapter 3: Radian Measure and the Unit Circle Approach, and Chapter 4: Graphing Trigonometric Functions. The core continues with Chapter 5: Trigonometric Identities, Chapter 6: Solving Trigonometric Equations, and Chapter 7: Applications of Trigonometry: Triangles and Vectors. The manual concludes with Chapter 8: Complex Numbers, Polar Coordinates, and Parametric Equations.

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Trigonometry – 5th Edition

SOLUTIONS
ST
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MANUAL
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Cynthia Y. Young
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Comprehensive Solution Manual for
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Instructors and Students
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© Cynthia Y. Young
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All rights reserved. Reproduction or distribution without permission is prohibited.




©MedConnoisseur

, Solution Manual for Trigonometry, 5th Edition by Cynthia Y. Young
CHAPTER 1
ST
Section 1.1 Solutions --------------------------------------------------------------------------------
1 x 1 x
1. Solve for x:  2. Solve for x: 
360∘ 360∘
UV
2 4
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∘
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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.)
_A
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∘ .
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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∘ .
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(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.
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a) complement: 90∘ 18∘  72∘ a) complement: 90∘  39∘  51∘
b) supplement: 180∘ 18∘  162∘ b) supplement: 180∘  39∘  141∘

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



1

, Chapter 1


13. 14.
a) complement: 90∘  89∘  1∘ a) complement: 90∘  75∘  15∘
ST
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
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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
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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
_A
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∘ .
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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
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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∘ .
– – – –
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 150∘  155∘



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

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


2

, Section 1.1



23.   180 ∘  53.3∘  23.6 ∘   103.1∘ 24.   180 ∘  105.6 ∘ 13.2∘   61.2 ∘
ST
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.
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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
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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.
_A
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 .
PP
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 .
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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
VE
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
D?
simplifies to a2  56, so we conclude that a  56  2 14 .




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