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MrSalmon: I think if you're going for that method then you have to use the [url= http://en.wikipedia.org/wiki/Law_of_cosines ]Law of Cosines[/url] first to get the angles.
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Do you need the angles? If you have split your triangle into two right angled triangles by dropping a perpendicular line from C in GrahamS diag to the side AB. Then you don't know the length of this line, say h. Likewise you dont know how far along it intersects along side AB, assume the distance is x from A. Using Pythagoras, you know b^2 = x^2 + h^2 and you also know a^2= (c-x)^2 + h^2, solve for h and x and work out the areas and add together.
On you go then...
mefty - for help, read the link in inigo's post.
The algebraic manipulation is stepped through very nicely in there.
If my algebra is right, and it has been a while since I've had to do stuff like this, then that should work as it should simplify to
2x^2+2cx+a^2-b^2-c^2 = 0.
You will need to solve the resultant quadratic equation which granted isn't as easy as that other formula but still doesn't require the use of any trig.