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I'm facing a problem. We have lengths of 3 segments. How to see if the triangle built of our 3 segments has a specific angle, for example 60°?

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up vote 1 down vote accepted

The law of cosines says that $$c^2=a^2+b^2-2ab\cos\gamma.$$ Thus $\gamma=60^\circ$ iff $$c^2=a^2+b^2-ab$$ (or with the sides permuted for the other angles).

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Thanks ;) I've forgotten that helpful law ;) – enedil Feb 6 '14 at 21:32

To find each specific angle, use the law of cosines:

$$c^2 = a^2 + b^2 - 2ab \cos C$$

Where $a, b, c$ are side lengths and $A, B, C$ are the angles opposite their corresponding sides.

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