If I remember correctly this is high school material; I feel ashamed that I can't solve this now.

Lengths of a triangle's sides determine its angles; but how to compute these angles?

enter image description here

  • $\begingroup$ I would recommend the Law of Cosines, since you don't know any of the angles. Solve as follows: $\dfrac{a^2+b^2-c^2}{2ab}=\cos(\gamma),$ where $\gamma$ is the angle opposite $c$. $\endgroup$ – Adrian Keister Jan 11 at 16:02

No, @ajotatxe, it's not the law of sine, it's rather the law of cosine $$c^2=a^2+b^2-2\ a\ b\ \cos(\gamma)$$ (which clearly can be solved for $cos(\gamma)$, and thus for $\gamma$.

--- rk


I'm not a massive fan of the use of $\gamma$ in the question, comment & answer.

To me, with the following image, although it is poorly drawn, is much easier to remember the cosine rule. It is a high-school formula after all!

We have that:

$$A^2=B^2+C^2-2BC\cos(a)\to a=\arccos\bigg(\frac{B^2+C^2-A^2}{2BC}\bigg)$$ $$B^2=A^2+C^2-2AC\cos(b)\to b=\arccos\bigg(\frac{A^2+C^2-B^2}{2AC}\bigg)$$ $$C^2=A^2+B^2-2AB\cos(c)\to c=\arccos\bigg(\frac{A^2+B^2-C^2}{2AB}\bigg)$$ enter image description here

  • 1
    $\begingroup$ In some countries are vertices denoted by capital letters, segments or lines by small letters of Latin alphabet, and angles by Greek alphabet. Moreover, side opposite to the vertex A is a, and angle at A is $\alpha.$ It is really practical. If you don't like, you've right to take other notation, but it is not a good idea to use the letters from OP in a different sense in the solution/answer. $\endgroup$ – user376343 Jan 12 at 11:54

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