If the sides of a triangle are p, q and sqrt{ | vedclass

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(B) Let the sides be $a = p$,$b = q$,and $c = \sqrt{p^2 + pq + q^2}$.Since $p, q > 0$,it is clear that $c^2 = p^2 + pq + q^2 > p^2$ and $c^2 > q^2$,so

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$2\pi / 3$

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(B) Let the sides be $a = p$,$b = q$,and $c = \sqrt{p^2 + pq + q^2}$.Since $p, q > 0$,it is clear that $c^2 = p^2 + pq + q^2 > p^2$ and $c^2 > q^2$,so $c$ is the largest side.The largest angle $	heta$ is opposite to the largest side $c$.Using the Law of Cosines: $\cos 	heta = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the values: $\cos 	heta = \frac{p^2 + q^2 - (p^2 + pq + q^2)}{2pq}$.$\cos 	heta = \frac{-pq}{2pq} = -\frac{1}{2}$.Since $\cos 	heta = -\frac{1}{2}$,we have $	heta = \arccos(-1/2) = \frac{2\pi}{3}$.

If the sides of a triangle are $p, q$ and $\sqrt{p^2 + pq + q^2}$,then the largest angle is

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