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1. Area and the Law of Sines

   Section 14.4

2. Area of a Triangle B c a h A C b 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 absin C orK = ½ acsin B.
3. Finding the Area 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: 2) a = 1.2 ft b = 0.9 ft mC = 33° 1) a = 10.1 m c = 9.8 m mB = 87° 3) b = 1 in c = 5 in mA = 20° Area = 49.4 m2 Area = 0.3 ft2 Area = 0.9 in2
4. The Law of Sines 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. Solving a Triangle B Round all answers to the nearest tenth. D 38 25 50° E 30° 44 108° F A C D = 100° e = 34.2 f = 22.3 A = 38.7° B = 33.3° b = 21.9
6. Assignment Complete Page 789 #1 - 16
7. For Tomorrow Complete: Page 793