Question

In the figure, O is the centre of the circle and $\angle POR={120}^o$. Find $\angle PQR$ and $\angle PSR$

Answer 1

Given, $\angle POR={120}^o$

Hence, $\angle POR=2\angle PQR$ (half angle property)

$\angle PQR=\frac{1}{2}\times {120}^o={60}^o$

$PQRS$ is a cyclic quadrilateral

$\therefore \angle PQR+\angle PSR={180}^o\angle PSR=180-60\angle PSR={120}^o$