# The resultant of two forces acting at any angle?

I am studying about forces as vectors. And they give me this equation:
$c^2 = a^2 + b^2 - 2ab \cos C$

Can anybody explain me the second part of the equation? I perfectly understand $c^2 = a^2 + b^2$ but what's with the $-2ab \,\mathrm{cos}\, C$?

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I fixed the formatting in two different ways, for the heck of it (go look). The first way is preferred for standard functions. –  Peter Morgan Jun 15 '11 at 17:43
@Peter Morgan, ah thanks! –  Dan the Man Jun 15 '11 at 17:54
Usage of this formula is needed when the basis of vector field is not orthogonal. Karthesian one ([1,0];[0,1]) is orthonormal, so you can use Pythagorean theorem, but in basis of [1,0];[1,1] you have to use law of cosines. –  Crowley Jun 16 '11 at 7:05
This is the law of cosines! The Pythagorean theorem is just a special case of this more general result - in the Pythagorean theorem we are dealing with a right triangle only, and $C=90$ degrees. This equation is true in the more general case of any triangle.