If the radius of the circumcircle of an isosceles triangle $P Q R$ is equal to $P Q (= P R)$ , then the angle $P$ , is

π / 6

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π / 3

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π / 2

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2 π / 3

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Solution

The correct option is C $2 π / 3$
In $Δ P Q R$
$P Q = P R = R_{1}$ and $Q = R$
where $R_{1}$ is circumradius
from sine rule: $\frac{P Q}{sin R} = 2 R_{1}$
$\Rightarrow sin R = \frac{1}{2}$
$\Rightarrow R = \frac{π}{6}$
Therefore, $P = π - Q - R = π - 2 R = \frac{2 π}{3}$
Ans: D

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