SSC CPO SI and ASI 2015 Paper

(3) Let PQ be the chord, KP and KQbe the radii with K as centre.

SinceKP PQ $=\mathit{P}\mathit{Q}=\mathit{K}\mathit{Q}$ , triangle KPQ is an equilateral. So, ∠ PKQ 60°
As angle made by an arc at centreis half of the angle made by it at circumference, therefore,
$\angle \mathit{P}\mathit{S}\mathit{Q}={30}^{°}$
since, $\mathrm{}\mathit{P}\mathit{S}\mathit{O}\mathit{R}$ is a cyclic quadrilateral, $\angle \mathit{P}\mathit{S}\mathit{Q}+\mathrm{angle}\mathrm{mathrm}\mathit{P}\mathit{R}\mathit{Q}={180}^{°}$
${30}^{°}+\angle \mathit{P}\mathit{R}\mathit{Q}={180}^{°}\Rightarrow \angle \mathit{P}\mathit{R}\mathit{Q}={150}^{°}$