SOLUTION: i have to work out the two solutions of the angle of two isosceles triangles with areas of 4cm squared and sides of 4.7 (no base length given) i know how to work out the area but

Question 590347: i have to work out the two solutions of the angle of two isosceles triangles with areas of 4cm squared and sides of 4.7 (no base length given)
i know how to work out the area but i cannot rearrange the equation 1/2ab x SinC to help me answer this question, please explain and give the two answers
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
(1/2)bh = 4 ___ h = 8 / b

b^2 + h^2 = 4.7^2

substituting ___ b^2 + (64 / b^2) = 4.7^2

(b^2)^2 + 64 = 22.09 b^2 ___ (b^2)^2 - 22.09 b^2 + 64 = 0

use quadratic formula to find b^2, then find b (two of the four solutions will not work)

cos(C) = (b/2) / 4.7
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