PQR is an isosceles triangle with PQ = PR. If the radius of the circum circle of $△PQR$ is equal to the length of PQ, then $\angle P=$

$\frac{QR}{\sin P}=\frac{PR}{\sin Q}=\frac{PQ}{\sin R}=2a$

Where a is the circum radius

given $PQ=PR=a=K$

$\begin{array}{r}\Rightarrow \frac{\mathrm{K}}{\sin \mathrm{Q}}=\frac{\mathrm{K}}{\sin \mathrm{R}}=2\mathrm{K}\\ \therefore \sin \mathrm{Q}=\sin \mathrm{R}\end{array}$

And

$\sin \mathrm{Q}=\sin \mathrm{R}=\frac{\mathrm{K}}{2\mathrm{K}}=\frac{1}{2}$

$\begin{array}{r}\therefore \mathrm{Q}=\mathrm{R}={30}^{\circ }\\ \therefore \mathrm{P}={180}^{\circ }-\left({30}^{\circ }+{30}^{\circ }\right)\\ ={120}^{\circ }=\frac{2\pi }{3}\end{array}$