2. a = 6, b = 4, C = 60 º 6 Sin A = 4 Sin B = c Sin 60º

3. a = 4, b = 5, c = 3 4 Sin A = 5 Sin B = 3 Sin C

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

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

6.  Solve these three equations for: Cos A Cos B = b + c - a 2bc 22 2 a + c - b 2ac 22 2 = a + b - c 2ab 22 2 Cos C=

7.  → Find the largest angle first a + c - b 2ac 22 2 Cos B= = 8 + 14 - 19 2(8)(14) 222 B = 116.8º → remember to use the inverse Cosine to find an angle

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

9. a) a = 5; b = 8; c = 9 b) a = 9; b = 7; c = 10

10. a + b - c 2ab 22 2 Cos C= = 5 + 8 - 9 2(5)(8) 2 22 C = 84.3º 5 Sin A 9 Sin 84.3º = A = 33.6º B = 62.2º

11. b + c - a 2bc 22 2 Cos A= = 7 + 9 - 10 2(7)(9) 2 22 A = 76.2º 10 Sin 76.2º 7 Sin B = B = 42.8º C = 61º

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

13.  Now use the Law of Sines 21.3 Sin 115º = 15 Sin B = 10 Sin C B = 39.7º C = 25.2º a = 21.3

14. a) a = 6; b = 4; C = 60 º b) a = 3; c = 2; B = 110 º

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

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

17. 5 8 c d 45º β d 2 = 5+ 8 – 2(5) (8) Cos 45º 22 d = 5.69

18. 5 8 c d 45º β 8 135º c 2 = 5+ 8 – 2(5) (8) Cos 135º 22 d = 12.07

19. 25 35 c d θ 120º c 2 = 25+ 35 – 2(25) (35) Cos 120º 22 c = 52.2 35

20. 25 35 c d 60º 8 120º d 2 = 25+ 35 – 2(25) (35) Cos 60º 22 d = 31.2 θ

21. Heron’s Area Formula

22.  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 = 43b = 53c = 72 s = Area = 84 (84 – 43) (84 – 53) (84 – 72) 43 + 53 + 72 2 = 84 = 1131.89 square meters

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

25.  A radio tower 500 feet high is located on the side of a hill with an inclination to the horizontal of 5 º. 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? 5º5º 100