2. A Triangle is uniquely determined by a “Contained Angle” A B O C b c a

3. A B O C b c a IF b,c and O are all known, then O is called a “Contained Angle”

4. A B O C b c a The Cosine Law can be used to find the length of the opposite side to O

5. A B 35 o C 5 8 a = ? Consider the following Triangle. Find a

6. A B 35 o C 5 X a = ? Construct a vertical line CN to create 2 right triangles N8 - X Label the height “h”, and the bases in terms of X h 8

7. A B 35 o C 5 X a = ? By the Pythagorean Theorm: N8 - X 5 2 = h 2 + X 2 h and a 2 = h 2 + (8 – X) 2

8. By the Pythagorean Theorm: 5 2 = h 2 + X 2 and a 2 = h 2 + (8 - X) 2 5 2 - X 2 = h 2 and h 2 = a 2 - (8 - X) 2 Since both equations are solved for h 5 2 - X 2 = a 2 - (8 - X) 2 a 2 = 5 2 - X 2 + (8 - X) 2 + (8 - X) 2 = + (8 - X) 2 - X 2 - (8 - X) 2

9. a 2 = 5 2 - X 2 + (8 - X) 2 a 2 = 5 2 - X 2 + 64 – 16X + X 2 a 2 = 25 + 64 – 16X From the Triangle, redefine X A 5 X 35 o Cos 35 o = X 5 X = 5Cos35 o

10. In the interest of efficiency, we hunt for a pattern that will allow us to generate a formula….take a few steps back: a 2 = 25 + 64 – 16X a 2 = 25 + 64 – 16(5Cos35 o ) a 2 = 25 + 64 – 80Cos35 o a 2 = 23.4678 a = 4.8 (side length opposite A)

11. If we can create a formula based on the steps, we can save a great amount of work. Examine the information given, and the structure of the final equation….

12. B a 2 = 25 + 64 – 80Cos35 o A 35 o C 5 8 a = ?

13. In General: a 2 = b 2 + c 2 – 2bcCosO o A B O C b c a

14. For Example: Find a a 2 = b 2 + c 2 – 2bcCosO o A 50 o C 8m 10m a B

15. a 2 = 8 2 + 10 2 – 2(8)(10)Cos50 o A 50 o C 8m 10m a a 2 = 61.15m a = 7.8 m B