6.2 Law of Cosines Objective Use the Law of Cosines to solve oblique triangles

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 (SSS and SAS) can be solved using the Law of Cosines. Law of Cosines Standard Form (SAS)Alternative Form (SSS)

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 Find the three angles of the triangle (SSS). Example 1: C BA Find the angle opposite the longest side first. Law of Sines: 36.3   26.4 

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 Solve the triangle (SAS). Example 2: 67.8  Law of Sines: 37.2  C BA  Law of Cosines:

Using the Law of Cosines to Solve a Triangle (SAS) Your Turn: Solve triangle ABC if A = 42.3°, b = 12.9 meters, and c = 15.4 meters. Solution Start by finding a using the law of cosines.

Using the Law of Cosines to Solve a Triangle (SAS)

Using the Law of Cosines to Solve a Triangle (SSS) Your Turn: Solve 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.

Using the Law of Cosines to Solve a Triangle (SSS) Verify with either the law of sines or the law of cosines that B  45.1°. Then,

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 Example 4 Application: 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

Example 3 – 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 (The pitcher’s mound is not halfway between home plate and second base.) How far is the pitcher’s mound from first base? Figure 6.13

Example 3 – 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 = – 2(43)(60) cos 45   So, the approximate distance from the pitcher’s mound to first base is  feet.

Assignment Pg 421 – 423; odd, 37, 39 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12