Trigonometry: A Unit Circle Approach 12th Edition – Solutions Manual (Michael Sullivan, 9780138234720) | Complete Answer Key

SOLUTIONS MANUAL

TRIGONOMETRY: A UNIT CIRCLE APPROACH
12TH EDITION


CHAPTER 1. GRAPHS AND FUNCTIONS
Section 1.1 (f) Quadrant IV

1. 0

2. 5   3  8  8

3. 32  42  25  5

4. 112  602  121  3600  3721  612
Since the sum of the squares of two of the sides
of the triangle equals the square of the third side,
the triangle is a right triangle.

1 16. (a) Quadrant I
5. bh
2 (b) Quadrant III
(c) Quadrant II
6. true (d) Quadrant I
7. x-coordinate or abscissa; y-coordinate or (e) y-axis
ordinate (f) x-axis

8. quadrants
9. midpoint
10. False; the distance between two points is never
negative.
11. False; points that lie in Quadrant IV will have a
positive x-coordinate and a negative y-coordinate.
The point  1, 4  lies in Quadrant II.

17. The points will be on a vertical line that is two
 x  x y  y2 
12. True; M   1 2 , 1 units to the right of the y-axis.
 2 2 

13. b
14. a
15. (a) Quadrant II
(b) x-axis
(c) Quadrant III
(d) Quadrant I
(e) y-axis

,18. The points will be on a horizontal line that is
three units above the x-axis. 28. d ( P1 , P2 )   6  ( 4) 2   2  (3) 2
 102  52  100  25
 125  5 5

29. d ( P1 , P2 )   2.3  (0.2) 2  1.1  (0.3) 2
 2.52  0.82  6.25  0.64
 6.89  2.62

30. d ( P1 , P2 )   0.3  1.2 2  1.1  2.32
19. d ( P1 , P2 )  (2  0) 2  (1  0) 2  (1.5) 2  (1.2) 2  2.25  1.44

 22  12  4  1  5  3.69  1.92


20. d ( P1 , P2 )  (2  0) 2  (1  0) 2 31. d ( P1 , P2 )  (0  a) 2  (0  b) 2

 (2) 2  12  4  1  5  (  a ) 2  ( b ) 2  a 2  b 2


21. d ( P1 , P2 )  (2  1) 2  (2  1) 2 32. d ( P1 , P2 )  (0  a ) 2  (0  a) 2

 (3) 2  12  9  1  10  (a )2  (a )2
 a 2  a 2  2a 2  a 2
22. d ( P1 , P2 )   2  (1)  2
 (2  1) 2

33. A  (2,5), B  (1,3), C  (1, 0)
 32  12  9  1  10
d ( A, B )  1  (2) 2  (3  5)2
23. d ( P1 , P2 )  (5  3) 2   4   4  
2
 32  (2) 2  9  4  13
 22   8   4  64  68  2 17  1  12  (0  3)2
2
d ( B, C ) 

 (2) 2  (3)2  4  9  13
 2   1    4  0 
2 2
24. d ( P1 , P2 ) 
d ( A, C )   1  (2) 2  (0  5)2
  3 2
 4  9  16  25  5
2

 12  (5) 2  1  25  26

25. d ( P1 , P2 )   4  (7) 2  (0  3)2
 112  ( 3) 2  121  9  130


26. d ( P1 , P2 )   4  2 2   2  (3) 2
 22  52  4  25  29

27. d ( P1 , P2 )  (6  5) 2  1  (2) 
2


 12  32  1  9  10

, Verifying that ∆ ABC is a right triangle by the  d ( A, B)2   d ( B, C )2   d ( A, C )2
Pythagorean Theorem:
10 2   10 2 
2 2
 d ( A, B)2   d ( B, C )2   d ( A, C )2   20 
2



 13    13     200  200  400
2 2 2
26
400  400
13  13  26 1
26  26 The area of a triangle is A  bh . In this
2
1 problem,
The area of a triangle is A   bh . In this
2
A    d ( A, B )    d ( B, C ) 
1
problem, 2
A  1   d ( A, B)    d ( B, C )  1
 10 2 10 2
2 2
 1  13  13  1 13 1
 100  2  100 square units
2 2 2
13
 2 square units
35. A  ( 5,3), B  (6, 0), C  (5,5)

34. A  (2, 5), B  (12, 3), C  (10,  11) d ( A, B)   6  ( 5) 2  (0  3)2
d ( A, B )  12  (2) 2  (3  5)2  112  ( 3) 2  121  9

 142  (2) 2  130

 196  4  200 d ( B, C )   5  6 2  (5  0)2
 10 2  (1) 2  52  1  25
d ( B, C )  10  12 2  (11  3)2  26

 (2)  (14)
2 2
d ( A, C )   5  ( 5) 2  (5  3)2
 4  196  200  102  22  100  4
 10 2  104
d ( A, C )  10  (2)  2
 (11  5) 2
 2 26

 122  (16) 2
 144  256  400
 20




Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:




Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:

,  d ( A, C )2   d ( B, C )2   d ( A, B)2  d ( A, C )2   d ( B, C )2   d ( A, B)2
         2   
2 2 2 2 2 2
104 26 130 29 29 145
104  26  130 29  4  29  145
130  130 29  116  145
1
The area of a triangle is A  bh . In this 145  145
2 1
problem, The area of a triangle is A  bh . In this
2
A    d ( A, C )    d ( B, C ) 
1 problem,
2
A    d ( A, C )    d ( B, C ) 
1
1
  104  26 2
2 1
1   29  2 29
  2 26  26 2
2 1
1   2  29
  2  26 2
2  29 square units
 26 square units
37. A  (4, 3), B  (0, 3), C  (4, 2)
36. A  (6, 3), B  (3, 5), C  (1, 5)
d ( A, B)  (0  4) 2   3  (3) 
2

d ( A, B)   3  (6)  2
 (5  3) 2

 ( 4)2  02  16  0
 92  (8) 2  81  64
 16
 145
4
d ( B, C )   1  32  (5  (5))2 d ( B, C )   4  0 2   2  (3) 2
 (4) 2  102  16  100
 42  52  16  25
 116  2 29
 41
d ( A, C )   1  ( 6)  2
 (5  3) 2
d ( A, C )  (4  4) 2   2  (3) 
2


 52  22  25  4
 02  52  0  25
 29
 25
5




Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem:
Verifying that ∆ ABC is a right triangle by the
Pythagorean Theorem: