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Section 1.6 Law of Cosines. Objectives: 1.To prove the law of cosines. 2.To solve triangles using the law of cosines. Law of Cosines For any triangle ABC, where side lengths opposite angles A, B, and C are a, b, and c respectively, then a 2 = b 2 + c 2 – 2bc cos A. - PowerPoint PPT Presentation
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Section 1.6Law of Cosines
Section 1.6Law of Cosines
Objectives:1. To prove the law of cosines.2. To solve triangles using the law
of cosines.
Objectives:1. To prove the law of cosines.2. To solve triangles using the law
of cosines.
Law of CosinesFor any triangle ABC, where side lengths opposite angles A, B, and C are a, b, and c respectively, then
a2 = b2 + c2 – 2bc cos A.
Law of CosinesFor any triangle ABC, where side lengths opposite angles A, B, and C are a, b, and c respectively, then
a2 = b2 + c2 – 2bc cos A.
Alternate forms of the Law of Cosinesa2 = b2 + c2 – 2bc cos Ab2 = a2 + c2 – 2ac cos Bc2 = a2 + b2 – 2ab cos C
Alternate forms of the Law of Cosinesa2 = b2 + c2 – 2bc cos Ab2 = a2 + c2 – 2ac cos Bc2 = a2 + b2 – 2ab cos C
To apply the Law of Cosines, you must know either the measures of all three sides or the measures of two sides and the included angle.
To apply the Law of Cosines, you must know either the measures of all three sides or the measures of two sides and the included angle.
EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.
f2 = d2 + g2 – 2dg cos F82 = 42 + 102 – 2(4)(10) cos F64 = 16 + 100 – 80 cos F64 = 116 – 80 cos F-52 = -80 cos FmF = cos-1 (0.65) ≈ 49.46 (4928΄)
f2 = d2 + g2 – 2dg cos F82 = 42 + 102 – 2(4)(10) cos F64 = 16 + 100 – 80 cos F64 = 116 – 80 cos F-52 = -80 cos FmF = cos-1 (0.65) ≈ 49.46 (4928΄)
EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.
g2 = d2 + f2 – 2df cos G102 = 42 + 82 – 2(4)(8) cos G100 = 16 + 64 – 64 cos G100 = 80 – 64 cos G20 = -64 cos GmG = cos-1 (-0.3125) ≈ 108.21 (10813΄)
g2 = d2 + f2 – 2df cos G102 = 42 + 82 – 2(4)(8) cos G100 = 16 + 64 – 64 cos G100 = 80 – 64 cos G20 = -64 cos GmG = cos-1 (-0.3125) ≈ 108.21 (10813΄)
EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.
mD = 180 – (mF + mG)mD = 180 – (49.46 + 108.21)mD = 180 – 157.67mD ≈ 22.33 (2219΄)
mD = 180 – (mF + mG)mD = 180 – (49.46 + 108.21)mD = 180 – 157.67mD ≈ 22.33 (2219΄)
EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10.
mF = 4928΄ f = 8mD = 2219΄ d = 4mG = 10813΄ g = 10
mF = 4928΄ f = 8mD = 2219΄ d = 4mG = 10813΄ g = 10
EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.
a2 = b2 + c2 – 2bc cos Aa2 = 122 + 92 – 2(12)(9) cos 63a2 = 144 + 81 – 216(.4540)a2 = 225 – 98.06a ≈ 11.3
a2 = b2 + c2 – 2bc cos Aa2 = 122 + 92 – 2(12)(9) cos 63a2 = 144 + 81 – 216(.4540)a2 = 225 – 98.06a ≈ 11.3
EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.
Use the law of sines to solve for C since C must be the smallest angle.Use the law of sines to solve for C since C must be the smallest angle.
EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.
asin A
asin A
csin C
csin C
=
11.3sin 63
11.3sin 63
9sin C
9sin C
=
9(sin 63)11.3
9(sin 63)11.3
sin C = sin C =
mC = sin-1 (0.70965) ≈ 45.2
EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.
mB = 180 – (mA + mC)mB = 180 – (63 + 45.2)mB = 180 – 108.2mB ≈ 71.8
mB = 180 – (mA + mC)mB = 180 – (63 + 45.2)mB = 180 – 108.2mB ≈ 71.8
EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9.
mA = 63 a = 11.3mB = 71.8 b = 12mC = 45.2 c = 9
mA = 63 a = 11.3mB = 71.8 b = 12mC = 45.2 c = 9
Practice: Find the mA if a = 38, b = 48, and c = 68. Round to the nearest degree.
Practice: Find the mA if a = 38, b = 48, and c = 68. Round to the nearest degree.
a2 = b2 + c2 – 2bc cos A382 = 482 + 682 – 2(48)(68) cos A1444 = 2304 + 4624 – 6528 cos A-5484 = -6528 cos AmA = cos-1 (0.8401) ≈ 33
a2 = b2 + c2 – 2bc cos A382 = 482 + 682 – 2(48)(68) cos A1444 = 2304 + 4624 – 6528 cos A-5484 = -6528 cos AmA = cos-1 (0.8401) ≈ 33
Practice: Find the mC if a = 38, b = 48, and c = 68. Round to the nearest degree.
Practice: Find the mC if a = 38, b = 48, and c = 68. Round to the nearest degree.
c2 = a2 + b2 – 2ab cos C682 = 382 + 482 – 2(38)(48) cos C4624 = 1444 + 2304 – 3648 cos C876 = -3648 cos CmC = cos-1 (-0.2401) ≈ 104
c2 = a2 + b2 – 2ab cos C682 = 382 + 482 – 2(38)(48) cos C4624 = 1444 + 2304 – 3648 cos C876 = -3648 cos CmC = cos-1 (-0.2401) ≈ 104
Homework
pp. 32-34
Homework
pp. 32-34
►A. Exercises►A. Exercises
1. a = 6, b = 5, c = 81. a = 6, b = 5, c = 862 = 52 + 82 – 2(5)(8)(cos A)36 = 25 + 64 – 80 cos A36 = 89 – 80 cos A
-53 = -80 cos Acos A = 0.6625mA = cos-1 (0.6625)mA ≈ 48.5°
62 = 52 + 82 – 2(5)(8)(cos A)36 = 25 + 64 – 80 cos A36 = 89 – 80 cos A
-53 = -80 cos Acos A = 0.6625mA = cos-1 (0.6625)mA ≈ 48.5°
52 = 62 + 82 – 2(6)(8)(cos B)25 = 36 + 64 – 96 cos B25 = 100 – 96 cos B
-75 = -96 cos Bcos B = 0.78125mB = cos-1 (0.78125)mB ≈ 38.6°
52 = 62 + 82 – 2(6)(8)(cos B)25 = 36 + 64 – 96 cos B25 = 100 – 96 cos B
-75 = -96 cos Bcos B = 0.78125mB = cos-1 (0.78125)mB ≈ 38.6°
►A. Exercises►A. Exercises
1. a = 6, b = 5, c = 81. a = 6, b = 5, c = 8
mC = 180° - mA - mBmC = 180° - 48.5° - 38.6°mC ≈ 92.9°
mC = 180° - mA - mBmC = 180° - 48.5° - 38.6°mC ≈ 92.9°
►A. Exercises►A. Exercises
1. a = 6, b = 5, c = 81. a = 6, b = 5, c = 8
a = 6
b = 5
c = 8
mA ≈ 48.5mB ≈ 38.6mC ≈ 92.9
a = 6
b = 5
c = 8
mA ≈ 48.5mB ≈ 38.6mC ≈ 92.9
►A. Exercises►A. Exercises
1.1.
►A. Exercises►A. Exercises
3. b = 26, c = 18, mA = 64°3. b = 26, c = 18, mA = 64°a2 = 262 + 182 – 2(26)(18)(cos 64°)a2 = 676 + 324 – 936(0.4384)a2 = 1000 – 410.3a2 = 589.7a ≈ 24.3
a2 = 262 + 182 – 2(26)(18)(cos 64°)a2 = 676 + 324 – 936(0.4384)a2 = 1000 – 410.3a2 = 589.7a ≈ 24.3
262 = 24.32 + 182 – 2(24.3)(18)(cos B)676 = 589.7 + 324 – 874.2(cos B)676 = 913.7 – 874.2(cos B)
-237.7 = -874.2(cos B)cos B = 0.2719mB = cos-1 (0.2719)mB ≈ 74.2°
262 = 24.32 + 182 – 2(24.3)(18)(cos B)676 = 589.7 + 324 – 874.2(cos B)676 = 913.7 – 874.2(cos B)
-237.7 = -874.2(cos B)cos B = 0.2719mB = cos-1 (0.2719)mB ≈ 74.2°
►A. Exercises►A. Exercises
3. b = 26, c = 18, mA = 64°3. b = 26, c = 18, mA = 64°
mC = 180° - mA - mBmC = 180° - 64° - 74.2°mC ≈ 41.8°
mC = 180° - mA - mBmC = 180° - 64° - 74.2°mC ≈ 41.8°
►A. Exercises►A. Exercises
3. b = 26, c = 18, mA = 64°3. b = 26, c = 18, mA = 64°
a ≈ 24.3
b = 26
c = 18
mA = 64mB ≈ 74.2mC ≈ 41.8
a ≈ 24.3
b = 26
c = 18
mA = 64mB ≈ 74.2mC ≈ 41.8
►A. Exercises►A. Exercises
3.3.
►A. Exercises►A. Exercises
7. A = 19.5°, B = 92°, c = 281. Basic trig ratios2. Law of sines3. Law of cosines
7. A = 19.5°, B = 92°, c = 281. Basic trig ratios2. Law of sines3. Law of cosines
►A. Exercises►A. Exercises
9. A = 60°, B = 90°, b = 101. Basic trig ratios2. Law of sines3. Law of cosines
9. A = 60°, B = 90°, b = 101. Basic trig ratios2. Law of sines3. Law of cosines
►B. Exercises►B. Exercises
11. A radio antenna is placed on the top of a 200-foot office building. The angle of elevation from a parking lot to the top of the antenna is 21°. The angle of depression looking from the bottom of the antenna to the lot is 10°. What is the height of the antenna?
11. A radio antenna is placed on the top of a 200-foot office building. The angle of elevation from a parking lot to the top of the antenna is 21°. The angle of depression looking from the bottom of the antenna to the lot is 10°. What is the height of the antenna?
►B. Exercises►B. Exercises
11.11.
200 ft200 ft
21°
10°
►B. Exercises►B. Exercises
11.11.
200 ft200 ft
69°
100°
11°
x
►B. Exercises►B. Exercises
11.11.
200 ft200 ft
10°
x
sin10
200x =
sin10 =200
x
x ≈ 1151.75
►B. Exercises►B. Exercises
11.11.
200 ft200 ft
69°
100°
11°
1151.75
x ≈ 235.4sin69
1151.75(sin11)x =
sin69
1151.75=
sin11
x
xx
■ Cumulative Review21. Convert 5 radians to degrees.■ Cumulative Review21. Convert 5 radians to degrees.
■ Cumulative Review22. Give the reference angle for -470°.■ Cumulative Review22. Give the reference angle for -470°.
■ Cumulative Review23. Write 3 reciprocal ratios.■ Cumulative Review23. Write 3 reciprocal ratios.
■ Cumulative Review24. In ∆ABC, find b if B = 27°, a = 8,
and A = 90°
■ Cumulative Review24. In ∆ABC, find b if B = 27°, a = 8,
and A = 90°
■ Cumulative Review25. In ∆ABC, find b if B = 27°, a = 8,
and A = 20°
■ Cumulative Review25. In ∆ABC, find b if B = 27°, a = 8,
and A = 20°
