Area of a TriangleDate: 10/18/2003 at 04:29:10 From: Kevin Subject: The Area of a Triangle Why can the area of a triangle be calculated using the formula K = a^2*sin(B)*sin(C)/[2*sin(B+C)] ? Date: 10/18/2003 at 11:17:05 From: Doctor Luis Subject: Re: The Area of a Triangle Hi Kevin, Please look at the following triangle. The area K of that triangle is K = (1/2)h*b So, to find the area, we need to find h and b. From trigonometry, we know that h = a sin(C). We can get b from the law of sines: sin(B)/b = sin(A)/a b = a*sin(B)/sin(A) Therefore, K = (1/2)(a*sin(C))*(a*sin(B)/sin(A)) = a^2 sin(B)sin(C)/(2*sin(A)) We're almost there. If you remember from geometry, the internal angles of a triangle sum to 180 degrees, or pi radians. A + B + C = pi A = pi - (B + C) Also, sin(A) = sin(pi - (B + C)) = sin(pi)cos(B+C) - cos(pi)sin(B+C) = 0*cos(B+C) - (-1)*sin(B+C) = sin(B + C) And that pretty much proves it: K = a^2*sin(B)*sin(C)/[2*sin(B + C)] I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
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