Pre-Calc Exam Notes 56

Pre-Calc Exam Notes 56 - − a 2 2 bc = 2 bc − b 2 − c...

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56 Chapter 2 General Triangles §2.4 In Figure 2.4.2, let D be the point where the altitude touches AB (or its extension). h b a A C B c D (a) A acute h b a A C B c D (b) A obtuse Figure 2.4.2 Proof of Heron’s formula By the Pythagorean Theorem, we see that h 2 = b 2 ( AD ) 2 . In Figure 2.4.2(a), we see that AD = b cos A . And in Figure 2.4.2(b) we see that AD = b cos (180 A ) = − b cos A . Hence, in either case we have ( AD ) 2 = b 2 (cos A ) 2 , and so h 2 = b 2 b 2 (cos A ) 2 = b 2 (1 (cos A ) 2 ) = b 2 (1 + cos A )(1 cos A ) . (2.31) (Note that the above equation also holds when A = 90 since cos 90 = 0 and h = b ). Thus, substituting equation (2.31) into equation (2.30), we have K 2 = 1 4 b 2 c 2 (1 + cos A )(1 cos A ) . (2.32) By the Law of Cosines we know that 1 + cos A = 1 + b 2 + c 2 a 2 2 bc = 2 bc + b 2 + c 2 a 2 2 bc = ( b + c ) 2 a 2 2 bc = (( b + c ) + a )(( b + c ) a ) 2 bc = ( a + b + c )( b + c a ) 2 bc , and similarly 1 cos A = 1 b 2 + c 2
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Unformatted text preview: − a 2 2 bc = 2 bc − b 2 − c 2 + a 2 2 bc = a 2 − ( b − c ) 2 2 bc = ( a − ( b − c ))( a + ( b − c )) 2 bc = ( a − b + c )( a + b − c ) 2 bc . Thus, substituting these expressions into equation (2.32), we have K 2 = 1 4 b 2 c 2 ( a + b + c )( b + c − a ) 2 bc · ( a − b + c )( a + b − c ) 2 bc = a + b + c 2 · b + c − a 2 · a − b + c 2 · a + b − c 2 , and since we de±ned s = 1 2 ( a + b + c ), we see that K 2 = s ( s − a )( s − b )( s − c ) , so upon taking square roots we get K = r s ( s − a )( s − b )( s − c ) . QED...
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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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