Question

The sides of a triangles are $a,b,\sqrt{a^2+ab+b^2}$,
prove that the greatest angles is ${120}^o$.

Answer 1

The side are $a,b,c$ where $c=\surd (a^2+b^2+ab)$ is greater side.

Hence $\angle C$ is greatest.

$\therefore \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{ab}{2ab}=\frac{1}{2}$

$\therefore C={120}^o$.