Name ________________________________________ Date __________________ Class__________________
LESSON
10-3
Reading Strategy
Use a Visual Map
The unit circle shown at right has a radius of 1 unit. So the terminal side of
each angle has a length of 1 unit. As shown in the diagram below, that
corresponds to the hypotenuse of a right triangle. The ordered pair shows
the length of each side of the triangle. Notice the ordered pairs for the
points at 0°, 90°, 180°, and 270°.
On the circle, for every value of θ,
sin θ =
y y
= =y
r 1
cos θ =
x x
= =x
r 1
tan θ =
y
x
Use the unit circle to answer each question.
1. Express 360° in radians.
________________________
4.
2. What is sin 360°?
_________________________
3. What is cos 360°?
________________________
a. In which quadrant does the terminal side of a
150° angle lie?
____________________
b. Express 150° in radians.
____________________
c. What point does the terminal side of an angle of
150° pass through on the unit circle?
____________________
d. Which coordinate of the ordered pair represents
sin 150°?
____________________
e. Which coordinate of the ordered pair represents
cos 150°?
____________________
f. Write an expression for tan 150° and simplify.
____________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
10-26
Holt McDougal Algebra 2
15.
3
16.
17.
2
2
;−
; −1
2
2
18.
1 3
3
;−
19. − ;
2 2
3
2
2
;−
;1
2
2
21. −
7. 30°
1
3
3
;−
;−
2
2
3
9. 45°
3 1
; ; 3
2 2
20. −
2
2
2
cos 45° =
2
10. sin 45° =
3 1
;− ; 3
2
2
tan 45° = 1
22. 2073 mi
2
2
2
cos 315° =
2
tan 315° = −1
11. sin 315° = −
Practice C
5π
radians
2
1. −270°
2.
3. 50°
4. −
5. 315°
6. −330°
35π
7.
radians
18
8. 63°
9.
π
radians
15
10π
radians
9
12. −
7π
radians
12
1 3
3
13. − ;
;−
2 2
3
14. −
2 2
;
; −1
2 2
15. −
17. −
19.
3 1
;− ; 3
2
2
16.
2
2
;−
;1
2
2
18.
2 2
;
;1
2 2
1
3
3
;−
21. − ; −
2
2
3
23. −
3 1
; ;− 3
2 2
Challenge
1. 6080 ft
2. 1,600,921 mi; 66,705 mi/h
3. Area of circle = πr 2; A sector whose central
angle has a measure of θ radians has an
θ
times the area of the circle. So
2π
θ
1
Area of sector =
πr 2 = θr 2 .
2π
2
10. 234°
37π
radians
30
11.
area of
( )
4.
3 1
;− ;− 3
2
2
b. θ =
2 2
;
; −1
2 2
3.
4
radians
7π
radians
6
5. 240°
2.
2π
π
or
6
3
π π
⋅
2 3
=
π2
6
e. Yes; possible answer: because the arc
length of the fragment is very close to
the arc length that would be expected
for a plate of diameter π
2
2
;−
; −1
2
2
2.
5π
radians
6
4. −
2
d. 1.64 in.
Reteach
π
π
c. S = r θ =
25. 138 ft
1. −
4
1. a. r =
20. 0; −1; 0
24.
π
Problem Solving
1
3
3
;−
;−
2
2
3
22. −
8. 300°
2π
radians
3
1
4
3. C
4. H
5. B
6. F
Reading Strategy
6. −270°
1. 2π
2. 0
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A35
Holt McDougal Algebra 2
3. 1
14. 71.8°
4. a. Quadrant II
16. 11° north of east
b.
5π
6
Practice B
⎛
3 1⎞
,
c. ⎜ −
⎜ 2 2 ⎟⎟
⎝
⎠
d.
1
2
3
2
1
2
−
3
2
=−
3
3
4.
5.
6.
π
4
π
3
3π
5π
+ 2π n;
+ 2π n
4
4
7π
+ 2π n
6
5π
; 300°
3
π
3
7.
3π
; 270°
2
9. 0; 0°
11π
; 330°
6
; 60°
11.
; 45°
13. 26.7°
16. 109.5°
17. 259.6°
18. 95.7°
19. 71°
+ 2π n;
5π
+ 2π n
3
+ 2π n;
5π
+ 2π n
4
+ 2π n;
2π
+ 2π n
3
7π
; 315°
4
π
4
Practice C
1.
2.
π
3
; 60°
3
π
4
+ 2π n;
2π
+ 2π n
3
+ 2π n;
7π
+ 2π n
4
2π
5π
+ 2π n;
+ 2π n
3
3
4.
3π
5π
+ 2π n;
+ 2π n
4
4
5.
9.
π
3.
π
6
+ 2π n;
5π
+ 2π n
6
6.
3π
7π
+ 2π n;
+ 2π n
4
4
5π
; 300°
3
; 30°
11.
7π
; 315°
4
7.
12. π; 180°
13.
3π
; 270°
2
9.
6
6
+ 2π n;
15. 247.5°
b. 0 radians, 0°
10.
π
14. 233.2°
2π
4π
+ 2π n;
+ 2π n
3
3
π
5.
5π
7π
+ 2π n;
+ 2π n
6
6
7. a. I and IV
8.
5π
7π
+ 2π n;
+ 2π n
4
4
12.
5π
+ 2π n;
+ 2π n
2.
6
6
3
4.
5π 7π
;
6 6
π
π
2π
4π
+ 2π n;
+ 2π n
3
3
10.
Practice A
3.
2.
8.
FUNCTIONS
b.
4π
5π
+ 2π n;
+ 2π n
3
3
6.
10-4 INVERSES OF TRIGONOMETRIC
1. a.
1.
3. 0 + 2π n; π + 2π n
e. −
f.
15. 281.5°
π
6
; 30°
8.
10.
2π
; 120°
3
π
2
; 90°
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A36
Holt McDougal Algebra 2