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6.2 Law of Cosines

6.2 Law of Cosines. What would happen if we were given:. a = 6, b = 4, C = 60 º. 6. 4. c. = . = . Sin A. Sin B. Sin 60 º. What would happen if we were given:. a = 4, b = 5, c = 3. 4. 5. 3. = . = . Sin A. Sin B. Sin C. Law of Cosines.

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6.2 Law of Cosines

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  1. 6.2 Law of Cosines

  2. What would happen if we were given: a = 6, b = 4, C = 60º 6 4 c = = Sin A Sin B Sin 60º

  3. What would happen if we were given: a = 4, b = 5, c = 3 4 5 3 = = Sin A Sin B Sin C

  4. Law of Cosines • Use Law of Cosines when the given information is: • SSS (three lowercase letters) • SAS (2 sides and an angle, all 3 letters)

  5. Law of Cosines 2 2 2 a b + c – 2bc Cos A = 2 2 2 b a + c – 2ac Cos B = 2 2 2 c a + b – 2ab Cos C =

  6. Alternative Form • Solve these three equations for: Cos A Cos B 2 2 2 2 2 2 a + b - c b + c - a = Cos C= 2ab 2bc 2 2 2 a + c - b = 2ac

  7. a = 8; b = 19; c = 14 (SSS) • → Find the largest angle first 2 2 2 2 2 2 a + c - b 8 + 14 - 19 = Cos B= 2ac 2(8)(14) → remember to use the inverse Cosine to find an angle B = 116.8º

  8. a = 8; b = 19; c = 14; B = 116.8º • Once you know one angle, it is easiest to now use the Law of Sines. 8 19 14 = = Sin A Sin 116.8º Sin C A = 22.08º C = 41.12º

  9. Solve the following triangles: • a = 5; b = 8; c = 9 • a = 9; b = 7; c = 10

  10. a = 5; b = 8; c = 9 C = 84.3º A = 33.6º B = 62.2º 2 2 2 2 2 2 a + b - c 5 + 8 - 9 = Cos C= 2ab 2(5)(8) 5 9 = Sin A Sin 84.3º

  11. a = 10; b = 7; c = 9 A = 76.2º B = 42.8º C = 61º 2 2 2 2 2 2 b + c - a 7 + 9 - 10 = Cos A= 2bc 2(7)(9) 10 7 = Sin 76.2º Sin B

  12. A = 115º; b = 15; c = 10 (SAS) • What do we have enough information to solve for? 2 2 2 a b + c – 2bc Cos A = 2 2 2 a = + 10 – 2(15) (10) Cos 115º 15 2 a a = = 451.78548 21.3

  13. a A = 115º; b = 15; c = 10 = 21.3 • Now use the Law of Sines 21.3 15 10 = = Sin B Sin C Sin 115º B = 39.7º C = 25.2º

  14. Solve the following triangles: • a = 6; b = 4; C = 60º • a = 3; c = 2; B = 110º

  15. a = 6; b = 4; C = 60º c = 5.29 B = 40.9º A = 79.1º 2 2 2 c a + b – 2ab Cos C = 4 5.29 = Sin 60º Sin B

  16. a = 3; c = 2; B = 110º b = 4.14 C = 27º A = 43º 2 2 2 b a + c – 2ac Cos B = 2 4.14 = Sin 110º Sin C

  17. Find the missing information: β c 5 d 45º 8 2 2 2 5 + 8 – 2(5) (8) Cos 45º d = d = 5.69

  18. Find the missing information: 8 β 135º c 5 d 45º 8 2 2 2 5 + 8 – 2(5) (8) Cos 135º c = d = 12.07

  19. Find the missing information: 35 120º c 25 d θ 35 2 2 2 25 + 35 – 2(25) (35) Cos 120º c = c = 52.2

  20. Find the missing information: 8 120º c 25 d 60º θ 35 2 2 2 25 + 35 – 2(25) (35) Cos 60º d = d = 31.2

  21. 6.2 Law of Cosines Heron’s Area Formula

  22. Heron’s Area Formula • Any triangle with given sides of lengths a, b, and c, has an area of: Area = s (s – a) (s – b) (s – c) where s = a + b + c 2

  23. Find the area of a triangle have sides of lengths a = 43 meters, b = 53 meters, and c = 72 meters. a = 43 b = 53 c = 72 s = Area = 84 (84 – 43) (84 – 53) (84 – 72) 43 + 53 + 72 = 84 2 = 1131.89 square meters

  24. 7 in. 5 in. a = 5 b = 7 c = 10 s = Area = 11 (11 – 5) (11 – 7) (11 – 10) 10 in. 5 + 7 + 10 = 11 2 = 16.25 square inches

  25. A radio tower 500 feet high is located on the side of a hill with an inclination to the horizontal of 5º. How long should two guy wires be if they are to connect to the top of the tower and be secured at two points 100 feet directly above and directly below the base of the tower? 100 100 5º

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