PQR is an isosceles triangle with PQ=PR. If t | vedclass

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(D) Let the circumradius be $R_c$. Given $R_c = PQ$. In $	riangle PQR$,$PQ = PR$,so $\angle Q = \angle R$. Using the sine rule,$\frac{PQ}{\sin R} = 2

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Let the circumradius be $R_c$. Given $R_c = PQ$. In $	riangle PQR$,$PQ = PR$,so $\angle Q = \angle R$. Using the sine rule,$\frac{PQ}{\sin R} = 2R_c$. Substituting $R_c = PQ$,we get $\frac{PQ}{\sin R} = 2PQ$. This implies $\sin R = \frac{1}{2}$,so $R = 30^{\circ}$. Since $\angle Q = \angle R$,we have $\angle Q = 30^{\circ}$. In $	riangle PQR$,the sum of angles is $180^{\circ}$,so $\angle P + \angle Q + \angle R = 180^{\circ}$. $\angle P + 30^{\circ} + 30^{\circ} = 180^{\circ} \Rightarrow \angle P = 120^{\circ}$.

$PQR$ is an isosceles triangle with $PQ=PR$. If the radius of the circumcircle of $\triangle PQR$ is equal to the length of $PQ$,then $\angle P=$ (in $^{\circ}$)

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