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Section 1.6 Law of Cosines

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.

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Section 1.6 Law of Cosines

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  1. Section 1.6 Law of Cosines

  2. Objectives: 1. To prove the law of cosines. 2. To solve triangles using the law of cosines.

  3. Law of Cosines For 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.

  4. Alternate forms of the Law of Cosines a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C

  5. 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.

  6. EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10. f2 = d2 + g2 – 2dg cos F 82 = 42 + 102 – 2(4)(10) cos F 64 = 16 + 100 – 80 cos F 64 = 116 – 80 cos F -52 = -80 cos F mF = cos-1 (0.65) ≈ 49.46 (4928΄)

  7. EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10. g2 = d2 + f2 – 2df cos G 102 = 42 + 82 – 2(4)(8) cos G 100 = 16 + 64 – 64 cos G 100 = 80 – 64 cos G 20 = -64 cos G mG = cos-1 (-0.3125) ≈ 108.21 (10813΄)

  8. EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10. mD = 180 – (mF + mG) mD = 180 – (49.46 + 108.21) mD = 180 – 157.67 mD ≈ 22.33 (2219΄)

  9. EXAMPLE 1 Solve FDG if f = 8, d = 4, and g = 10. mF = 4928΄ f = 8 mD = 2219΄ d = 4 mG = 10813΄ g = 10

  10. EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9. a2 = b2 + c2 – 2bc cos A a2 = 122 + 92 – 2(12)(9) cos 63 a2 = 144 + 81 – 216(.4540) a2 = 225 – 98.06 a ≈ 11.3

  11. 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.

  12. a sin A c sin C 9 sin C = 11.3 sin 63 = 9(sin 63) 11.3 sin C = EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9. mC = sin-1 (0.70965) ≈ 45.2

  13. EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9. mB = 180 – (mA + mC) mB = 180 – (63 + 45.2) mB = 180 – 108.2 mB ≈ 71.8

  14. EXAMPLE 2 Solve ABC if A = 63, b = 12, and c = 9. mA = 63 a = 11.3 mB = 71.8 b = 12 mC = 45.2 c = 9

  15. Practice: Find the mA if a = 38, b = 48, and c = 68. Round to the nearest degree. a2 = b2 + c2 – 2bc cos A 382 = 482 + 682 – 2(48)(68) cos A 1444 = 2304 + 4624 – 6528 cos A -5484 = -6528 cos A mA = cos-1 (0.8401) ≈ 33

  16. Practice: Find the mC if a = 38, b = 48, and c = 68. Round to the nearest degree. c2 = a2 + b2 – 2ab cos C 682 = 382 + 482 – 2(38)(48) cos C 4624 = 1444 + 2304 – 3648 cos C 876 = -3648 cos C mC = cos-1 (-0.2401) ≈ 104

  17. Homework pp. 32-34

  18. ►A. Exercises 1. a = 6, b = 5, c = 8 62 = 52 + 82 – 2(5)(8)(cos A) 36 = 25 + 64 – 80 cos A 36 = 89 – 80 cos A -53 = -80 cos A cos A = 0.6625 mA = cos-1 (0.6625) mA ≈ 48.5°

  19. ►A. Exercises 1. a = 6, b = 5, c = 8 52 = 62 + 82 – 2(6)(8)(cos B) 25 = 36 + 64 – 96 cos B 25 = 100 – 96 cos B -75 = -96 cos B cos B = 0.78125 mB = cos-1 (0.78125) mB ≈ 38.6°

  20. ►A. Exercises 1. a = 6, b = 5, c = 8 mC = 180° - mA - mB mC = 180° - 48.5° - 38.6° mC ≈ 92.9°

  21. ►A. Exercises 1. a = 6 b = 5 c = 8 mA ≈ 48.5 mB ≈ 38.6 mC ≈ 92.9

  22. ►A. Exercises 3. b = 26, c = 18, mA = 64° a2 = 262 + 182 – 2(26)(18)(cos 64°) a2 = 676 + 324 – 936(0.4384) a2 = 1000 – 410.3 a2 = 589.7 a ≈ 24.3

  23. ►A. Exercises 3. b = 26, c = 18, mA = 64° 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 = 0.2719 mB = cos-1 (0.2719) mB ≈ 74.2°

  24. ►A. Exercises 3. b = 26, c = 18, mA = 64° mC = 180° - mA - mB mC = 180° - 64° - 74.2° mC ≈ 41.8°

  25. ►A. Exercises 3. a ≈24.3 b = 26 c = 18 mA = 64 mB ≈ 74.2 mC ≈ 41.8

  26. ►A. Exercises 7. A = 19.5°, B = 92°, c = 28 1. Basic trig ratios 2. Law of sines 3. Law of cosines

  27. ►A. Exercises 9. A = 60°, B = 90°, b = 10 1. Basic trig ratios 2. Law of sines 3. Law of cosines

  28. ►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?

  29. ►B. Exercises 11. 10° 200 ft 21°

  30. ►B. Exercises 11. 69° 100° x 200 ft 11°

  31. 200 sin10 = x 200 x = sin10 ►B. Exercises 11. x ≈ 1151.75 x 200 ft 10°

  32. 1151.75 x = sin69 sin11 1151.75(sin11) x = sin69 ►B. Exercises 11. 69° x 100° x ≈ 235.4 1151.75 200 ft 11°

  33. ■ Cumulative Review 21. Convert 5 radians to degrees.

  34. ■ Cumulative Review 22. Give the reference angle for -470°.

  35. ■ Cumulative Review 23. Write 3 reciprocal ratios.

  36. ■ Cumulative Review 24. In ∆ABC, find b if B = 27°, a = 8, and A = 90°

  37. ■ Cumulative Review 25. In ∆ABC, find b if B = 27°, a = 8, and A = 20°

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