Area of a Triangle
Date: 6/2/96 at 19:53:54 From: Anonymous Subject: Area of a triangle Hello, How can I calculate the area of a triangle if I know that one of its sides measures l and that its two adjacent angles measure a and b? I have tried many things, and I feel that I should use trigonometry, though I am not sure how. Maybe I should think about a parallelogram and the congruent angles formed by its diagonals?
Date: 6/3/96 at 9:33:39 From: Doctor Pete Subject: Area of a triangle Say you had /|\ / | \ / | \ / |h \ / | \ a /_____|__________\ b x l-x so the sum of the base is x+l-x = l, your given length. Then tan(a) = h/x, and tan(b) = h/(l-x), so h = x tan(a) = (l-x) tan(b) ==> x(tan(a)+tan(b)) = l tan(b) ==> x = l tan(b)/(tan(a)+tan(b)). Hence h = x tan(a) = l*tan(a)tan(b)/(tan(a)+tan(b)). Then the area of the triangle is: h*l l^2*tan(a)tan(b) --- = ---------------- . 2 2(tan(a)+tan(b)) Alternatively, this can be rewritten to include the third angle c, which is Pi-(a+b) radians or 180-(a+b) degrees: l^2 / 1 \ Area = --- | ------------- + cot(c) | , 2 \ tan(a)+tan(b) / which might be easier to use. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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