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

Solution manual
Solution
for trigonometry
manual for5th
trigonometry
edition by cynthia
5th edition
y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
by cynthia y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
Page 1https://online2pdf.com/remove-pdf-header-footer




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

Page 1 of 954https://online2pdf.com/remove-pdf-header-footer
Solution manual
1 of 954https://online2pdf.com/remove-pdf-header-footer
for trigonometry 5th edition by cynthia y younghttps://online2pdf.com/remove-pdf-header-footer

,Solution manual
Solution
for trigonometry
manual for5th
trigonometry
edition by cynthia
5th edition
y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
by cynthia y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
Page 2https://online2pdf.com/remove-pdf-header-footer


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∘ .


2

Page 2 of 954https://online2pdf.com/remove-pdf-header-footer
Solution manual
2 of 954https://online2pdf.com/remove-pdf-header-footer
for trigonometry 5th edition by cynthia y younghttps://online2pdf.com/remove-pdf-header-footer

,Solution manual
Solution
for trigonometry
manual for5th
trigonometry
edition by cynthia
5th edition
y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
by cynthia y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
Page 3https://online2pdf.com/remove-pdf-header-footer


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 .




3

Page 3 of 954https://online2pdf.com/remove-pdf-header-footer
Solution manual
3 of 954https://online2pdf.com/remove-pdf-header-footer
for trigonometry 5th edition by cynthia y younghttps://online2pdf.com/remove-pdf-header-footer

, Solution manual
Solution
for trigonometry
manual for5th
trigonometry
edition by cynthia
5th edition
y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
by cynthia y young.pdfhttps://online2pdf.com/remove-pdf-header-footer
Page 4https://online2pdf.com/remove-pdf-header-footer


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.



4

Page 4 of 954https://online2pdf.com/remove-pdf-header-footer
Solution manual
4 of 954https://online2pdf.com/remove-pdf-header-footer
for trigonometry 5th edition by cynthia y younghttps://online2pdf.com/remove-pdf-header-footer