Question

In the given figure, PQRS is a cyclic quadrilateral in a circle with centre O. If $\angle PSR={130}^{\circ }$. Find $\angle QPR$.​​​​​​

Answer 1

Given, $\angle PSR={130}^{\circ }$
We have $\angle PSR+\angle PQR={180}^{\circ }$ [Opposite angles of a cyclic quadrilateral]
$\Rightarrow {130}^{\circ }+\angle Q={180}^{\circ }$
$\Rightarrow \angle Q={180}^{\circ }-{130}^{\circ }={50}^{\circ }$
Also, $\angle PRQ={90}^{\circ }$ [Angle subtended by diameter on the circle is always ${90}^{\circ }$]
In $\Delta PRQ,$
$\angle QPR+\angle PRQ+\angle RQP={180}^{\circ }$ [Sum of angles of triangle is ${180}^{\circ }$]
$\Rightarrow \angle QPR+{90}^{\circ }+{50}^{\circ }={180}^{\circ }$
$\Rightarrow \angle QPR={180}^{\circ }-{140}^{\circ }={40}^{\circ }$