Question

'O' is center of circle $C_1$ and AB is the diameter of circle $C_2$. Quadrilateral PQRS is inscribed in the cirlce with center 'O'. Find $\angle QRS$

Answer 1

R.E.F image

$\angle R_1=\angle ORS$

$\angle R_2=\angle ORQ$

$\angle AOB={90}^{\circ }$ (Property of angle subtended by

diameter AB)

So, OS = OR = OQ = radius ; Then

${△}^{\prime }sSOR$ and $ROQ$ are isosceles.

Let $\angle S=\angle R_1=\angle x$

and $R_2=\angle Q=y$

In Quadrilateral SORQ,

$(\angle x+\angle y)\times 2+{90}^{\circ }={360}^{\circ }$

So, $\angle R=\angle x+\angle y={135}^{\circ }$