Name ________________________________________ Date __________________ Class__________________
LESSON
8-2
Reteach
Trigonometric Ratios
Trigonometric Ratios
sin A =
leg opposite ∠A 4
= = 0.8
hypotenuse
5
hypotenuse
leg opposite ∠A
leg adjacent to ∠A 3
= = 0.6
hypotenuse
5
leg opposite ∠A
4
tan A =
= ≈ 1.33
leg adjacent to ∠A 3
cos A =
leg adjacent to ∠A
You can use special right triangles to write trigonometric ratios as fractions.
sin 45° = sin Q =
=
leg opposite ∠Q
hypotenuse
x
x 2
=
So sin 45° =
=
1
2
2
2
2
.
2
Write each trigonometric ratio as a fraction and as a decimal
rounded to the nearest hundredth.
1. sin K
2. cos H
_________________________________________
3. cos K
________________________________________
4. tan H
_________________________________________
________________________________________
Use a special right triangle to write each trigonometric ratio as a
fraction.
5. cos 45°
6. tan 45°
_________________________________________
7. sin 60°
________________________________________
8. tan 30°
_________________________________________
________________________________________
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8-14
Holt McDougal Geometry
Name ________________________________________ Date __________________ Class__________________
Reteach
LESSON
8-2
Trigonometric Ratios continued
You can use a calculator to find the value of trigonometric ratios.
cos 38° ≈ 0.7880107536 or about 0.79
You can use trigonometric ratios to find side lengths of triangles.
Find WY.
cos W =
adjacent leg
hypotenuse
cos 39° =
7.5 cm
WY
Substitute the given values.
WY =
7.5
cos 39°
Solve for WY.
WY ≈ 9.65 cm
Write a trigonometric ratio that involves WY.
Simplify the expression.
Use your calculator to find each trigonometric ratio. Round to the
nearest hundredth.
9. sin 42°
10. cos 89°
_________________________________________
11. tan 55°
________________________________________
12. sin 6°
_________________________________________
________________________________________
Find each length. Round to the nearest hundredth.
13. DE
14. FH
_________________________________________
15. JK
________________________________________
16. US
_________________________________________
________________________________________
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8-15
Holt McDougal Geometry
1
base × height.
2
1
Substitution gives Area = bc sin A.
2
8-2 TRIGONOMETRIC RATIOS
a triangle is Area =
Practice A
a b
;
c
c
1. opposite; hypotenuse;
2. 18.66 m2
4. 12.05 in2
b a
2. adjacent; hypotenuse;
;
c
c
3. opposite; adjacent;
a
b
;
a
3
; 0.60
5
5.
6.
4
; 1.33
3
7. 0.54
8. 0.68
5. Possible answer: The Pythagorean
Theorem shows that x2 + y2 = c2. It also
shows that (b − x)2 + y2 = a2. Expanding
the latter equation gives b2 − 2bx + x2 + y2
= a2. Substituting, b2 − 2bx + c2 = a2. But
x
cos A = , so x = c cos A. Another
c
substitution gives a2 = b2 + c2 − 2bc cos A.
b
4.
4
; 0.80
5
6. 6.78 cm
9. 6.31
10. 12.46 m
11. 19.70 mm
12. 2.55 feet
13. 277 feet
3. 28.39 ft2
7. 15.18 km
8. 22.83 ft
Reteach
Practice B
1.
15
≈ 0.88
17
2.
15
≈ 0.88
17
8
≈ 0.47
17
4.
8
≈ 0.53
15
1.
7
; 0.28
25
2.
7
; 0.28
25
3.
3.
24
; 3.43
7
4.
24
; 0.96
25
5.
2
2
6.
1
=1
1
5.
24
; 0.96
25
6.
7
; 0.29
24
7.
3
2
8.
3
3
7.
1
2
8.
3
2
9. 1
10.
12.
11.
2
2
9. 0.67
10. 0.02
11. 1.43
12. 0.10
3
3
13. 39.65 m
14. 6.01 in.
15. 32.91 mm
16. 55.32 cm
3
Challenge
13. 0.90
14. 0.53
15. 0.27
16. 14.03 in.
17. 57.36 cm
18. 0.36 mi
19. 8.68 km
20. 95.41 yd
1, 2, 3, 4, 5.
21. 3.18 ft
Practice C
1. Possible answer: Draw an altitude from
h
∠B and call its length h. Then sin A = ,
c
so h = c sin A. The formula for the area of
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A12
Holt McDougal Geometry