Question

If the sides of the triangle are $a,b$ and $\sqrt{a^2+ab+b^2}$,then find the greatest angle

Answer 1

Given sides are, $a,b,\sqrt{a^2+b^2+ab}$
So the greatest angle will be corresponding to greatest side which is $\sqrt{a^2+b^2+ab}$
Thus using cosine rule,
$\cos \theta =\frac{a^2+b^2-(a^2+b^2+ab)}{2ab}=-\frac{1}{2}\Rightarrow {\theta }_{max}={120}^0$