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$\mathrm{PQR}$ 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 $\angle \mathrm{P}=$
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Verified Answer
The correct answer is:
$120^{\circ}$
$$
R_c=P Q=\dot{b}=c
$$
Using sine rule
$$
\begin{aligned}
& \frac{P Q}{\sin R}=2 R_c=P Q \times 2 \\
& \sin R=\frac{1}{2}, R=30^{\circ} \\
& \because \angle Q=\angle R \Rightarrow Q=30^{\circ} \\
& \therefore P=120^{\circ} .
\end{aligned}
$$
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}
$$
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