SOLUTION: Hi i have a question on trigonometry that i am finding quite challenging: Prove that the triangle ABC can be found from: � c� Sin A Sin B Sin (A+B) its 1/2 c� SinA

Question 435666: Hi i have a question on trigonometry that i am finding quite challenging:

Prove that the triangle ABC can be found from:
� c� Sin A Sin B
Sin (A+B)
its 1/2 c� SinASinB/Sin(A+B)
i have tried using the three area of triangle formulas but the derivation is beyond me thanks
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Let A , B, and C be the interior angles of the triangle, and a, b, and c, their respective opposite sides.
Then the area of the triangle ABC is given by
.
Now from the sine law, ==> , and
==> . Then
.
But C = 180 - (A+B)==> sinC = sin(180-(A+B)) = sin(A+B), whence
.

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