Pre-Calc Exam Notes 55

# Pre-Calc Exam Notes 55 - △ ABC with sides a b and c let s...

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The Area of a Triangle Section 2.4 55 Case 2: Three angles and any side. Suppose that we have a triangle ABC in which one side, say, a , and all three angles are known. 7 By the Law of Sines we know that c = a sin C sin A , so substituting this into formula (2.24) we get: Area = K = a 2 sin B sin C 2 sin A (2.26) Similar arguments for the sides b and c give us: Area = K = b 2 sin A sin C 2 sin B Area = K = c 2 sin A sin B 2 sin C (2.27) (2.28) Example 2.14 b a = 12 A = 115 C = 40 B = 25 c Find the area of the triangle ABC given A = 115 , B = 25 , C = 40 , and a = 12. Solution: Using formula (2.26), the area K is given by: K = a 2 sin B sin C 2 sin A = 12 2 sin 25 sin 40 2 sin 115 K = 21.58 Case 3: Three sides. Suppose that we have a triangle ABC in which all three sides are known. Then Heron’s formula 8 gives us the area: Heron’s formula: For a triangle
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Unformatted text preview: △ ABC with sides a , b , and c , let s = 1 2 ( a + b + c ) (i.e. 2 s = a + b + c is the perimeter of the triangle). Then the area K of the triangle is Area = K = r s ( s − a )( s − b )( s − c ) . (2.29) To prove this, ±rst remember that the area K is one-half the base times the height. Using c as the base and the altitude h as the height, as before in Figure 2.4.1, we have K = 1 2 hc . Squaring both sides gives us K 2 = 1 4 h 2 c 2 . (2.30) 7 Note that this is equivalent to knowing just two angles and a side (why?). 8 Due to the ancient Greek engineer and mathematician Heron of Alexandria (c. 10-70 A.D.)....
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## This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.

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