In the given figure, PQRS is a cyclic quadrilateral in a circle with centre O. If ∠PSR=130∘. Find ∠QPR.
40∘
Given, ∠PSR=130∘
We have ∠PSR+∠PQR=180∘ [Opposite angles of a cyclic quadrilateral]
⇒130∘+∠Q=180∘
⇒∠Q=180∘−130∘=50∘
Also, ∠PRQ=90∘ [Angle subtended by diameter on the circle is always 90∘]
In ΔPRQ,
∠QPR+∠PRQ+∠RQP=180∘ [Sum of angles of triangle is 180∘]
⇒∠QPR+90∘+50∘=180∘
⇒∠QPR=180∘−140∘=40∘