Question

If the radius of the circumcircle of an isosceles triangle $PQR$ is equal to PQ which is equal to PR, then the $\angle$P

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is D $\frac{2\pi }{3}$

Sine rule $:$

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

where $a$ is the circumradius

Given $:$ $PQ=PR=a=K\left(let\right)$

$\Rightarrow \frac{K}{\sin Q}=\frac{K}{\sin R}=2K\therefore \sin Q=\sin R$

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

$\therefore Q=R=30°$

$\therefore P=180°-(30°+30°)$

$=120°=\frac{2\pi }{3}$

Answer D