Student Solution Manual for Trigonometry 5th Edition by Cynthia Young| 9781119880028| All Chapters| LATEST

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∘




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,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
1 17∘ –33∘    180∘ And So,   30∘ . 1 10∘ –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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,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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, 33. Since This Is A Right Triangle, We Know From The Pythagorean Theorem That

 7
2
A2  B2  C2. Using The Given Information, This Becomes A2   52 , 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 36. If X  8 M, Then The Hypotenuse
Hypotenuse Of This Triangle Has Of This Triangle Has Length 8 2  11.31
Length
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
Of This Triangle Has Length 10 Ft., Then 2 X  10, So
10 10
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.




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