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

• A

  $30^{\circ}$
• B

  $60^{\circ}$
• C

  $45^{\circ}$
• Correct answer
  D

  $120^{\circ}$

Solution

$$ R_c=P Q=\dot{b}=c $$ Using sine rule $$ \begin{aligned} & \frac{P Q}{\sin R}=2 R_c=P Q \times 2 \\ & \sin R=\frac{1}{2}, R=30^{\circ} \\ & \because \angle Q=\angle R \Rightarrow Q=30^{\circ} \\ & \therefore P=120^{\circ} . \end{aligned} $$

Hence, the correct answer is option: 'D'.