1. Law of Cosines 8.2

2. 2  Use the Law of Cosines to solve oblique triangles (SSS or SAS).  Use the Law of Cosines to model and solve real-life problems.  Use Heron’s Area Formula to find the area of a triangle. Objectives

3. 3 Introduction

4. 4 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 An oblique triangle is a triangle that has no right angles. To solve an oblique triangle, you need to know the measure of at least one side and the measures of any other two parts of the triangle – two sides, two angles, or one angle and one side. C BA a b c

5. 5 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 The following cases are considered when solving oblique triangles. 1.Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) A C c A B c a c b C c a c a B Introduction

6. 6 When you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete. In such cases, you can use the Law of Cosines. The last two cases (SSS and SAS) can be solved using the Law of Cosines. (The first two cases can be solved using the Law of Sines.)

7. 7 Derivation of the Law of Cosines Let ABC be any oblique triangle drawn with its vertices labeled as in the figure below. The coordinates of point A become (c cos B, c sin B). Figure 10 pg 10-28

8. 8 Derivation of the Law of Cosines Point C has coordinates (a, 0) and AC has length b. This result is one form of the law of cosines. Placing A or C at the origin would have given the same result, but with the variables rearranged.

9. 9 The Law of Cosines If A, B, and C are the measures of the angles of a triangle, and a, b, and c are the lengths of the sides opposite these angles, then a 2  b 2  c 2  2bc cos A b 2  a 2  c 2  2ac cos B c 2  a 2  b 2  2ab cos C. The square of a side of a triangle equals the sum of the squares of the other two sides minus twice their product times the cosine of their included angle.

10. 10 Law of Cosines

11. 11 LAW OF COSINES Use these to find missing sides Use these to find missing angles Law of Cosines

12. 12 Solving an SAS Triangle 1.Use the Law of Cosines to find the side opposite the given angle. 2.Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute. 3.Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180º.

13. 13 Step l Use the Law of Cosines to find the side opposite the given angle. Thus, we will find a. SolutionWe are given two sides and an included angle. Therefore, we apply the three  step procedure for solving a SAS triangle. a 2  20 2  30 2  2(20)(30) cos 60º b = 20, c = 30, and A = 60°.  400  900  1200(0.5)  700 Perform the indicated operations. a 2  b 2  c 2  2bc cos A Apply the Law of Cosines to find a. Take the square root of both sides and solve for a. a   700  26 a b = 20 c = 30 A B C 60º Example: SAS Triangle Solve the triangle shown with A = 60º, b = 20, and c = 30.

14. 14 a b = 20 c = 30 A B C 60º Find sin -1 0.6547 using a calculator. Example cont. Solve the triangle shown with A = 60º, b = 20, and c = 30. Step 2 Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute. The shorter of the two given sides is b  20. Thus, we will find acute angle B. Solution Apply the Law of Sines. We are given b = 20 and A = 60°. Use the exact value of a, ÷ 700, from step 1. Cross multiply. Divide by square root of 700 and solve for sin B.

15. 15 SolutionWe are given two sides and an included angle. Therefore, we apply the three  step procedure for solving a SAS triangle. Step 3 Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180º. C  180º  A  B  180º  60º  41º  79º The solution is a  26, B  41º, and C  79º. a b = 20 c = 30 A B C 60º Example cont. Solve the triangle shown with A = 60º, b = 20, and c = 30.

16. 16 Solve Δ ABC with a = 11, c = 14, and B = 34°. SOLUTION Use the law of cosines to find side length b. b 2 = a 2 + c 2 – 2ac cos B b 2 = 11 2 + 14 2 – 2(11)(14) cos 34° b 2 61.7 7.85 Law of cosines Substitute for a, c, and B. Simplify. Take positive square root. Your Turn:

17. 17 Use the law of sines to find the measure of angle A. sin A a sin B b = sin A 11 = sin 34° 7.85 sin A = 11 sin 34° 7.85 0.7836 A sin –1 0.7836 51.6° Law of sines Substitute for a, b, and B. Multiply each side by 11 and Simplify. Use inverse sine. The third angle C of the triangle is C 180° – 34° – 51.6° = 94.4°. In ABC, b 7.85, A 51.68, and C 94.48. ANSWER Your Turn cont.

18. 18 Example:Solve triangle ABC if A = 42.3°, b = 12.9 meters, and c = 15.4 meters. Your Turn:

19. 19 B must be the smaller of the two remaining angles since it is opposite the shorter of the two sides b and c. Therefore, it cannot be obtuse. Caution If we had chosen to find C rather than B, we would not have known whether C equals 81.7 ° or its supplement, 98.3 °. Your Turn cont.

20. 20 Solving an SSS Triangle 1.Use the Law of Cosines to find the angle opposite the longest side. 2.Use the Law of Sines to find either of the two remaining acute angles. 3.Find the third angle. Subtract the measures of the angles found in steps 1 and 2 from 180º.

21. 21 Example – SSS Triangle Find the three angles of the triangle shown below. Solution: It is a good idea first to find the angle opposite the longest side—side b in this case. Using the alternative form of the Law of Cosines, you find that

22. 22 Example – Solution Because cos B is negative, B is an obtuse angle given by B  116.80 . At this point, it is simpler to use the Law of Sines to determine A. cont’d

23. 23 Example – Solution Because B is obtuse and a triangle can have at most one obtuse angle, you know that A must be acute. So, A  22.08  and C  180  – 22.08  – 116.80  = 41.12 . cont’d

24. 24 Law of Cosines Do you see why it was wise to find the largest angle first in the Example? Knowing the cosine of an angle, you can determine whether the angle is acute or obtuse. That is, cos  > 0 for 0  <  < 90  cos  < 0 for 90  <  < 180 . So, in the Example, once you found that angle B was obtuse, you knew that angles A and C were both acute. Furthermore, if the largest angle is acute, then the remaining two angles are also acute. Acute Obtuse

25. 25 Solve ABC with a = 12, b = 27, and c = 20. SOLUTION First find the angle opposite the longest side, AC. Use the law of cosines to solve for B. b 2 = a 2 + c 2 – 2ac cos B 27 2 = 12 2 + 20 2 – 2(12)(20) cos B 27 2 = 12 2 + 20 2 – 2(12)(20) = cos B – 0.3854 cos B B cos –1 (– 0.3854) 112.7° Law of cosines Substitute. Solve for cos B. Simplify. Use inverse cosine. Your Turn:

26. 26 Now use the law of sines to find A. sin A a = sin B b sin A 12 sin 112.7° 27 = sin A= 12 sin 112.7° 27 0.4100 A sin –1 0.4100 24.2° Law of sines Substitute for a, b, and B. Multiply each side by 12 and simplify. Use inverse sine. The third angle C of the triangle is C 180° – 24.2° – 112.7° = 43.1°. In ABC, A 24.2, B 112.7, and C 43.1. ANSWER Your Turn cont.

27. 27 ExampleSolve triangle ABC if a = 9.47 feet, b =15.9 feet, and c = 21.1 feet. Solution We solve for C, the largest angle, first. If cos C < 0, then C will be obtuse. Your Turn:

28. 28 Verify with either the law of sines or the law of cosines that B  45.1°. Then, Your Turn cont.

29. 29 Applications

30. 30 Example – An Application of the Law of Cosines The pitcher’s mound on a women’s softball field is 43 feet from home plate and the distance between the bases is 60 feet, as shown in Figure 3.9. (The pitcher’s mound is not halfway between home plate and second base.) How far is the pitcher’s mound from first base? Figure 3.9

31. 31 Example – Solution In triangle HPF, H = 45  (line HP bisects the right angle at H), f = 43, and p = 60. Using the Law of Cosines for this SAS case, you have h 2 = f 2 + p 2 – 2fp cos H = 43 2 + 60 2 – 2(43)(60) cos 45   1800.3. So, the approximate distance from the pitcher’s mound to first base is  42.43 feet.

32. 32 SolutionAfter two hours. the plane flying at 325 miles per hour travels 325 · 2 miles, or 650 miles. Similarly, the plane flying at 300 miles per hour travels 600 miles. The situation is illustrated in the figure. Let b  the distance between the planes after two hours. We can use a north  south line to find angle B in triangle ABC. Thus, B  180º  66º  26º  88º. We now have a  650, c  600, and B  88º. Example Two airplanes leave an airport at the same time on different runways. One flies at a bearing of N66ºW at 325 miles per hour. The other airplane flies at a bearing of S26ºW at 300 miles per hour. How far apart will the airplanes be after two hours?

33. 33 b 2 = a 2 + c 2 - 2ac cos B Apply the Law of Cosines. SolutionWe use the Law of Cosines to find b in this SAS situation. b 2 = 650 2 + 600 2 - 2(650)(600) cos 88º Substitute: a= 650, c =600, and B= 88°. After two hours, the planes are approximately 869 miles apart. b = 869 Solve for b. Example cont.

34. 34 Scientists can use a set of footprints to calculate an organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to 180°, the more efficiently the organism walked. The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B. Your Turn:

35. 35 SOLUTION b 2 = a 2 + c 2 – 2ac cos B 316 2 = 155 2 + 197 2 – 2(155)(197) cos B 316 2 = 155 2 + 197 2 – 2(155)(197) = cos B – 0.6062 cos B B cos –1 (– 0.6062) 127.3° Use inverse cosine. Simplify. Solve for cos B. Substitute. Law of cosines The step angle B is about 127.3°. ANSWER Solution

36. 36 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 36 Two ships leave a port at 9 A.M. One travels at a bearing of N 53  W at 12 mph, and the other travels at a bearing of S 67  W at 16 mph. How far apart will the ships be at noon? 53  67  c 36 mi 48 mi C At noon, the ships have traveled for 3 hours. Angle C = 180  – 53  – 67  = 60  The ships will be approximately 43 miles apart. 43 mi 60  N Your Turn:

37. 37 Your Turn:

38. 38 Heron’s Area Formula

39. 39 Heron’s Area Formula The Law of Cosines can be used to establish the following formula for the area of a triangle. This formula is called Heron’s Area Formula after the Greek mathematician Heron.

40. 40 Example 5 – Using Heron’s Area Formula Find the area of a triangle having sides of lengths a = 43 meters, b = 53 meters, and c = 72 meters. Solution: Because s = (a + b + c)/2 = 168/2 = 84, Heron’s Area Formula yields

41. 41 Example 5 – Solution  1131.89 square meters. cont’d

42. 42 Heron’s Area Formula You have now studied three different formulas for the area of a triangle. Standard Formula: Area = bh Oblique Triangle: Area = bc sin A = ab sin C = ac sin B Heron’s Area Formula: Area =

43. 43 Trigonometry Pages 291-294 8, 10, 12, 14, 16, 18, 20, 24, 26, 28 Chapter 3-2 Homework Part 1