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RE: Math problem (triangle area solution)

To: "Adamson, Ken" <KADAMSON@lifeline.net>,
Subject: RE: Math problem (triangle area solution)
From: "Adamson, Ken" <KADAMSON@lifeline.net>
Date: Wed, 5 Nov 2003 11:12:58 -0600
> 10400 (10414)
> 
> s = (a+b+c)/2
>
> K = (s*(s-a)*(s-b)*(s-c))^(1/2)


Hmmm - my answer doesn't seem to jibe here...

I'll work it a different way then :)

Dang - I found the problem.  My algebra is a little rusty...

-8425 = -(10625 * CosC)                                 (do some algebra)

8425 = 10625 * CosC                                     (some more algebra)

should be

-8425 = -(10625 * CosC)                                 (do some algebra)

8425 = 10625 * -CosC                                    (some more algebra)

-CosC = 8425/10625

CosC = -8425/10625

C = 142.461 degrees

K = 1/2 abSinC

K = .5 * 125 * 170 * Sin(142.461)

K = 6473.793 square units                               (same answer...)

Funny thing about complementary angles, they have the same sine...

Ken
'01 PT Cruiser Limited
HS101 in OKC (Soon to be STS101)





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