If the radius of the circumcircle of an isosc | vedclass

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(D) Let $R$ be the circumradius of $\Delta PQR$. Given $R = PQ = PR$.By the sine rule,$R = \frac{PQ}{2 \sin R} = \frac{PR}{2 \sin Q} = \frac{QR}{2 \si

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$\frac{2\pi}{3}$

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(D) Let $R$ be the circumradius of $\Delta PQR$. Given $R = PQ = PR$.By the sine rule,$R = \frac{PQ}{2 \sin R} = \frac{PR}{2 \sin Q} = \frac{QR}{2 \sin P}$.Since $R = PQ$,we have $PQ = \frac{PQ}{2 \sin R}$,which implies $\sin R = \frac{1}{2}$.Thus,$\angle R = \frac{\pi}{6}$.Since $PQ = PR$,the triangle is isosceles,so $\angle Q = \angle R = \frac{\pi}{6}$.The sum of angles in a triangle is $\pi$. Therefore,$\angle P = \pi - (\angle Q + \angle R) = \pi - (\frac{\pi}{6} + \frac{\pi}{6}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

If the radius of the circumcircle of an isosceles triangle $PQR$ is equal to $PQ$ (where $PQ = PR$),then the angle $P$ is

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