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13-6

13-6. The Law of Cosines. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 2. Holt Algebra 2. Warm Up Find each measure to the nearest tenth. 1 . m  y 2. x 3. y 4. What is the area of ∆ XYZ ? Round to the nearest square unit. ≈ 8.8. 104°. ≈ 18.3.

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13-6

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  1. 13-6 The Law of Cosines Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

  2. Warm Up Find each measure to the nearest tenth. 1.my2. x 3. y 4. What is the area of ∆XYZ ? Round to the nearest square unit. ≈ 8.8 104° ≈ 18.3 60 square units

  3. Objectives Use the Law of Cosines to find the side lengths and angle measures of a triangle. Use Heron’s Formula to find the area of a triangle.

  4. In the previous lesson, you learned to solve triangles by using the Law of Sines. However, the Law of Sines cannot be used to solve triangles for which side-angle-side (SAS) or side-side-side (SSS) information is given. Instead, you must use the Law of Cosines.

  5. To derive the Law of Cosines, draw∆ABC with altitude BD. If x represents the length of AD, the length of DC is b – x.

  6. In ∆ABD, cos A = or x = cos A. Substitute c cos A for x. Write an equation that relates the side lengths of ∆DBC. a2 = (b–x)2 + h2 Pythagorean Theorem Expand (b – x)2. a2 = b2– 2bx + x2 + h2 In ∆ABD, c2 = x2 + h2. Substitute c2 for x2 + h2. a2 = b2– 2bx +c2 a2 = b2– 2b(c cos A)+ c2 a2 = b2 + c2– 2bccos A The previous equation is one of the formulas for the Law of Cosines.

  7. Example 1A: Using the Law of Cosines Use the given measurements to solve ∆ABC. Round to the nearest tenth. a = 8, b = 5, mC = 32.2° Step 1 Find the length of the third side. c2 = a2 + b2– 2ab cos C Law of Cosines c2 = 82 + 52– 2(8)(5)cos 32.2° Substitute. c2 ≈ 21.3 Use a calculator to simplify. c ≈ 4.6 Solve for the positive value of c.

  8. Solve for m B. Example 1A Continued Step 2 Find the measure of the smaller angle, B. Law of Sines Substitute. Solve for sin B.

  9. Solve for m A. Example 1A Continued Step 3 Find the third angle measure. mA + 35.4° + 32.2°  180° Triangle Sum Theorem mA 112.4°

  10. Step 1 Find the measure of the largest angle, B. m B = Cos-1 (0.2857) ≈ 73.4° Solve for m B. Example 1B: Using the Law of Cosines Use the given measurements to solve ∆ABC. Round to the nearest tenth. a = 8, b = 9, c = 7 b2 = a2 + c2– 2ac cos B Law of cosines 92 = 82 + 72– 2(8)(7)cos B Substitute. cos B = 0.2857 Solve for cos B.

  11. m C = Cos-1 (0.6667) ≈ 48.2° Solve for m C. Example 1B Continued Use the given measurements to solve ∆ABC. Round to the nearest tenth. Step 2 Find another angle measure c2 = a2 + b2– 2ab cos C Law of cosines 72 = 82 + 92– 2(8)(9)cos C Substitute. cos C = 0.6667 Solve for cos C.

  12. m A + 73.4° + 48.2°  180° m A 58.4° Solve for m A. Example 1B Continued Use the given measurements to solve ∆ABC. Round to the nearest tenth. Step 3 Find the third angle measure. Triangle Sum Theorem

  13. b = 23, c = 18, m A = 173° Check It Out! Example 1a Use the given measurements to solve ∆ABC. Round to the nearest tenth. Step 1 Find the length of the third side. a2 = b2 + c2– 2bc cos A Law of Cosines a2 = 232 + 182– 2(23)(18)cos 173° Substitute. a2 ≈ 1672.8 Use a calculator to simplify. a ≈ 40.9 Solve for the positive value of c.

  14. Solve for m C. m C = Sin-1 Check It Out! Example 1a Continued Step 2 Find the measure of the smaller angle, C. Law of Sines Substitute. Solve for sin C.

  15. m B + 3.1° + 173°  180° m B 3.9° Solve for m B. Check It Out! Example 1a Continued Step 3 Find the third angle measure. Triangle Sum Theorem

  16. Step 1 Find the measure of the largest angle, C. m C = Cos-1 (0.1560) ≈ 81.0° Solve for m C. Check It Out! Example 1b Use the given measurements to solve ∆ABC. Round to the nearest tenth. a = 35, b = 42, c = 50.3 c2 = a2 + b2– 2ab cos C Law of cosines 50.32 = 352 + 422– 2(35)(50.3) cos C Substitute. cos C = 0.1560 Solve for cos C.

  17. m A = Cos-1 (0.7264) ≈ 43.4° Solve for m A. Check It Out! Example 1b Continued Use the given measurements to solve ∆ABC. Round to the nearest tenth. a = 35, b = 42, c = 50.3 Step 2 Find another angle measure a2 = c2 + b2– 2cb cos A Law of cosines 352 = 50.32 + 422– 2(50.3)(42) cos A Substitute. cos A = 0.7264 Solve for cos A.

  18. m B + 81° + 43.4°  180° m B 55.6° Solve for m B. Check It Out! Example 1b Continued Step 3 Find the third angle measure.

  19. Remember! The largest angle of a triangle is the angle opposite the longest side.

  20. 1 Understand the Problem Example 2: Problem-Solving Application If a hiker travels at an average speed of 2.5 mi/h, how long will it take him to travel from the cave to the waterfall? Round to the nearest tenth of an hour. The answer will be the number of hours that the hiker takes to travel to the waterfall.

  21. 1 Understand the Problem The answer will be the number of hours that the hiker takes to travel to the waterfall. List the important information: • The cave is 3 mi from the cabin. • The waterfall is 4 mi from the cabin. The path from the cabin to the waterfall makes a 71.7° angle with the path from the cabin to the cave. • The hiker travels at an average speed of 2.5 mi/h.

  22. Make a Plan 2 Use the Law of Cosines to find the distance d between the water-fall and the cave. Then determine how long it will take the hiker to travel this distance.

  23. 3 Solve Step 1 Find the distance d between the waterfall and the cave. d2 = c2 + w2– 2cw cos D Law of Cosines Substitute 4 for c, 3 for w, and 71.7 for D. d2 = 42 + 32– 2(4)(3)cos71.7° d2 ≈ 17.5 Use a calculator to simplify. d ≈ 4.2 Solve for the positive value of d.

  24. The hiker must travel about 4.2 mi to reach the waterfall. At a speed of 2.5 mi/h, it will take the hiker ≈ 1.7 h to reach the waterfall. Look Back 4 Step 2 Determine the number of hours. In 1.7 h, the hiker would travel 1.7  2.5 = 4.25 mi. Using the Law of Cosines to solve for the angle gives 73.1°. Since this is close to the actual value, an answer of 1.7 hours seems reasonable.

  25. Check It Out! Example 2 A pilot is flying from Houston to Oklahoma City. To avoid a thunderstorm, the pilot flies 28° off the direct route for a distance of 175 miles. He then makes a turn and flies straight on to Oklahoma City. To the nearest mile, how much farther than the direct route was the route taken by the pilot?

  26. 1 Understand the Problem The answer will be the additional distance the pilot had to fly to reach Oklahoma City. List the important information: • The direct route is 396 miles. • The pilot flew for 175 miles off the course at an angle of 28° before turning towards Oklahoma City.

  27. Make a Plan 2 Use the Law of Cosines to find the distance from the turning point on to Oklahoma City. Then determine the difference additional distance and the direct route.

  28. 3 Solve Step 1 Find the distance between the turning point and Oklahoma City. Use side-angle-side. b2 = c2 + a2– 2ca cos B Law of Cosines b2 = 3962 + 1752–2(396)(175)cos28° Substitute 396 for c, 175 for a, and 28° for B. b2 ≈ 65072 Use a calculator to simplify. b ≈ 255 Solve for the positive value of b.

  29. Look Back 4 Step 2 Determine the number of additional miles the plane will fly. Add the actual miles flown and subtract from that normal distance to find the extra miles flown Total miles traveled. 255 + 175 = 430 430 – 396 = 34 Additional miles. By using the Law of Cosines the length of the extra leg of the trip could be determined.

  30. The Law of Cosines can be used to derive a formula for the area of a triangle based on its side lengths. This formula is called Heron’s Formula.

  31. Example 3: Landscaping Application A garden has a triangular flower bed with sides measuring 2 yd, 6 yd, and 7 yd. What is the area of the flower bed to the nearest tenth of a square yard? Step 1 Find the value of s. Use the formula for half of the perimeter. Substitute 2 for a, 6 for b, and 7 for c.

  32. A = A = Example 3 Continued Step 2 Find the area of the triangle. Heron’s formula Substitute 7.5 for s. A = 5.6 Use a calculator to simplify. The area of the flower bed is 5.6 yd2.

  33. m C ≈ 112.0° Solve for m c. Find the area of the triangle by using the formula area = ab sin c. area Example 3 Continued Check Find the measure of the largest angle, C. c2 = a2+ b2– 2ab cos C Law of Cosines 72 = 22+ 62– 2(2)(6) cos C Substitute. Solve for cos C. cos C ≈ –0.375 

  34. Check It Out! Example 3 The surface of a hotel swimming pool is shaped like a triangle with sides measuring 50 m, 28 m, and 30 m. What is the area of the pool’s surface to the nearest square meter? Step 1 Find the value of s. Use the formula for half of the perimeter. Substitute 50 for a, 28 for b, and 30 for c.

  35. A = Check It Out! Example 3 Continued Step 2 Find the area of the triangle. Heron’s formula Substitute 54 for s. Use a calculator to simplify. A = 367 m2 The area of the flower bed is 367 m2.

  36. m A ≈ 119.0° Solve for m A. Find the area of the triangle by using the formula area = ab sin A. Check It Out! Example 3 Continued Check Find the measure of the largest angle, A. 502 = c2+ b2– 2cb cos A Law of Cosines 502 = 302+ 282– 2(30)(28) cos A Substitute. Solve for cos A. cos A ≈ –0.4857 

  37. c ≈ 41.7; m A ≈ 25.4°; m B ≈ 72.1° m A ≈ 142.6°; m B ≈ 19.7°; m C ≈ 17.7° Lesson Quiz: Part I Use the given measurements to solve ∆ABC. Round to the nearest tenth. 1. a = 18, b = 40, m C = 82.5° 2. a = 18.0; b = 10; c = 9

  38. Lesson Quiz: Part II 3. Two model planes take off from the same spot. The first plane travels 300 ft due west before landing and the second plane travels 170 ft southeast before landing. To the nearest foot, how far apart are the planes when they land? 437 ft 4. An artist needs to know the area of a triangular piece of stained glass with sides measuring 9 cm, 7 cm, and 5 cm. What is the area to the nearest square centimeter? 17 cm2

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