1. 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. 60 square units 104° ≈ 8.8 ≈ 18.3

2. 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. Objectives

3. 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.

4. 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.

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

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. c 2 = a 2 + b 2 – 2ab cos C c 2 = 8 2 + 5 2 – 2(8)(5) cos 32.2° c 2 ≈ 21.3 c ≈ 4.6 Law of Cosines Substitute. Use a calculator to simplify. Solve for the positive value of c.

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

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

10. 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 Step 1 Find the measure of the largest angle, B. b 2 = a 2 + c 2 – 2ac cos B 9 2 = 8 2 + 7 2 – 2(8)(7) cos B cos B = 0.2857 m B = Cos -1 (0.2857) ≈ 73.4° Law of cosines Substitute. Solve for cos B. Solve for m B.

11. Example 1B Continued Use the given measurements to solve ∆ABC. Round to the nearest tenth. Step 2 Find another angle measure c 2 = a 2 + b 2 – 2ab cos C Law of cosines 7 2 = 8 2 + 9 2 – 2(8)(9) cos C Substitute. cos C = 0.6667 Solve for cos C. m C = Cos -1 (0.6667) ≈ 48.2° Solve for m C.

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

13. 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. a 2 = b 2 + c 2 – 2bc cos A a 2 = 23 2 + 18 2 – 2(23)(18) cos 173° a 2 ≈ 1672.8 a ≈ 40.9 Law of Cosines Substitute. Use a calculator to simplify. Solve for the positive value of c. b = 23, c = 18, m A = 173°

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

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

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

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

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

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

20. 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. 1 Understand the Problem The answer will be the number of hours that the hiker takes to travel to the waterfall.

21. 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. 1 Understand the Problem The answer will be the number of hours that the hiker takes to travel to the waterfall.

22. 2 Make a Plan 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. d 2 = c 2 + w 2 – 2cw cos D d 2 = 4 2 + 3 2 – 2(4)(3)cos 71.7° d 2 ≈ 17.5 d ≈ 4.2 Law of Cosines Substitute 4 for c, 3 for w, and 71.7 for D. Use a calculator to simplify. Solve for the positive value of d. Solve 3 Step 1 Find the distance d between the waterfall and the cave.

24. Step 2 Determine the number of hours. 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 Back4 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. 2 Make a Plan 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. b 2 = c 2 + a 2 – 2ca cos B b 2 = 396 2 + 175 2 – 2(396)(175)cos 28° b 2 ≈ 65072 b ≈ 255 Law of Cosines Substitute 396 for c, 175 for a, and 28° for B. Use a calculator to simplify. Solve for the positive value of b. Solve 3 Step 1 Find the distance between the turning point and Oklahoma City. Use side-angle-side.

29. 255 + 175 = 430 430 – 396 = 34 Total miles traveled. Additional miles. 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 Look Back4 By using the Law of Cosines the length of the extra leg of the trip could be determined.