Law of Cosines Objective The student will identify

Law of Cosines Objective: The student will identify, create, and solve practical problems involving triangles. Techniques will include using the trigonometric functions, The Pythagorean Theorem, the Law of Sines, and the Law of Cosines.

Law of Cosines • The Law of Sines can be used when we know one complete pair within a triangle • The size of an angle and the length of the side across from that angle. • If this is not the case, however, we need another method for solving non-right triangles. • This method is called the Law of Cosines.

Law of Cosines • Need three pieces of information in the triangle • Two sides and the angle between those sides (SAS) • All three sides (SSS) • When you have the option, always find the angles in order from largest to smallest if you use the Law of Cosines, or smallest to largest if you use the Law of Sines.

Law of Cosines • The Law of Cosines can take three different forms, depending upon which sides and/or angle you are given. • a 2 = b 2 + c 2 – 2 bc • cos A • b 2 = a 2 + c 2 – 2 ac • cos B • c 2 = a 2 + b 2 – 2 ab • cos C

Law of Cosines • The third version looks a lot like the Pythagorean Theorem. The Pythagorean Theorem is a special case of the Law of Cosines. c 2 = a 2 + b 2 – 2 ab • cos C c 2 = a 2 + b 2 – 2 ab • cos 90° c 2 = a 2 + b 2 – 2 ab • 0 c 2 = a 2 + b 2 – 0 c 2 = a 2 + b 2 Note: cos 90° = 0

Law of Cosines • If you know all three sides, and need to find the angles, the three formulas can be rewritten in terms of the angle as follows: • cos A = • cos B = • cos C =

Example 1: • Solve the triangle where a = 5, b = 7, and C = 43° B c A 5 7 43° 4. 7754 • sin A = 5 • sin 43° C c 2 = a 2 + b 2 – 2 ab • cos C c 2 = 52 + 72 – 2(5)(7) • cos 43° c 2 = 22. 8052 c = 4. 7754 c = 4. 8 B = 180 – 43 – 46 B = 91° sin A = 0. 71406 A = sin-1(0. 71406) A = 45. 56° A = 46°

Example 2: • Solve the triangle where a = 12, b = 17, and c = 23 B 23 A 12 17 C A = 180 – 104 – 46 A = 30° C = 103. 6089 C = 104° B = 45. 9207 B = 46°

Example 3: • A road goes up a hill for 895 feet, then flattens out by 8° and continues up the hill for another 542 feet. Engineers want to build a road that climbs the hill at a constant slope. What would be the length of the new C road? 542 ft 8° B 895 ft A 172° b b 2 = a 2 + c 2 – 2 ac • cos B b 2 = 5422 + 8952 – 2(542)(895) • cos 172° b 2 = 2, 055, 527. 275 b = 1433. 711015 b = 1434 ft