Section 14.5

1. Law of Cosines Section 14.5

2. Area and Pieces Find the area of the triangle shown. C 4 mi 65˚ A B 5 mi Area = ½ bc sin A = ½ (4) (5) sin 65˚ ≈ 9.06 mi2 Now, find the missing side and angles for the triangle. We do not have a method for a situation like this. We need a new strategy.

3. A New Formula C In ΔACD, b2 = x2 + h2 and cosA = x/b, or x = bcosA. a h b In ΔCBD, a2 = (c – x)2 + h2 c – x = c2 – 2cx + x2 + h2 x A B Using substitution we get = c2 – 2cx + b2 D Also, we can get = c2 – 2cbcosA + b2 With some minor rearrangement we get a formula called the Law of Cosines: a2 = b2 + c2 – 2bccosA b2 = a2 + c2 – 2accosB c2 = a2 + b2 – 2abcosC

4. Pieces of the Triangle Use the Law of Cosines to find the missing pieces of this triangle. C 4 mi 65˚ A B 5 mi b2 = a2 + c2 – 2accosB 42 = 4.92 + 52 – 2(4.9)(5) cosB 16 = 24.1 + 25 – 49 cosB 16 = 49.1 – 49 cosB –33.1 = –49 cosB 0.6755 ≈ cosB B = cos–1 (0.6755) = 47.5˚ C = 180˚ – A – B = 180˚ – 65˚ – 47.5˚ = 67.5˚ a2 = b2 + c2 – 2bccosA = 42 + 52 – 2(4)(5) cos 65˚ = 16 + 25 – 40 cos 65˚ ≈ 41 – 40(0.4226) ≈ 24.09526953 a ≈ 4.9 mi

5. Find x Use the Law of Cosines or the Law of Sines to find x. Round to 2 decimal places. 25˚ 10 x x 65˚ 13 32˚ 9 x = 12.61 x = 11.29

6. Homework For tomorrow: Page 796 #1 – 17