# 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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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.

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oh! i see! so the "c^2 = a^2 + b^2" is just a shorthand when its cosine is equal to 90, because cos 90 = 0. –  Daniel Pendergast Jun 15 '11 at 17:58
Well, not exactly ANY triangle, because there is Spherical triangle. ;) But when talking about planar geometry, yes, you are completely right. –  Crowley Jun 16 '11 at 7:01

Thanks for the question. You can derive this equation from the Pythagorean equation which is only for the the case when the two lines a and b are perpendicular to each other. But if they are at an angle then this is the general equation.

Now from the picture \begin{align} c^2&=(a+b \cos \alpha)^2 + (b \sin \alpha)^2\\ & = a^2 + 2ab \cos \alpha + b^2(\cos^2 \alpha + \sin^2 \alpha)\\ & = a^2 + b^2 + 2ab \cos \alpha \end{align}

Now cos alpha can be written as; $$cos (180 - C) = -cos C$$ Thus the equation becomes $$c^2 = a^2 + b^2 - 2ab cos C$$

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