1. Area and the Law of Sines

2. A B C a b c h The area, K, of a triangle is K = ½ bh where h is perpendicular to b (called the altitude). Using Right Triangle ratios we have sin A = h / c. If we solve for h by multiplying both sides by c we get h = c sin A. If we substitute this in the formula for area we know have K = ½ bc sin A. Similarly we could also have K = ½ ab sin C or K = ½ ac sin B.

3. We can now find the area of any triangle if we know two sides and the angle between them. Sketch a picture and find the area, to the nearest tenth, of the following triangles: 1) a = 10.1 m c = 9.8 m m  B = 87° 2) a = 1.2 ft b = 0.9 ft m  C = 33° 3) b = 1 in c = 5 in m  A = 20° Area = 49.4 m 2 Area = 0.3 ft 2 Area = 0.9 in 2

4. If we use the transitive property on our three area formulas we get: ½ bc sin A = ½ ac sin B = ½ ab sin C Dividing all terms by ½ abc gives us the Law of Sines: sin A a sin B b sin C c = = The Law of Sines can be used to solve any triangle if we have a side, the angle opposite it, and any other piece.

5. D E F 50° 30° 44 AC B 25 38 108°Round all answers to the nearest tenth. D = 100° e = 34.2 f = 22.3 A = 38.7° B = 33.3° b = 21.9

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