Question

In the figure above, $PQRS$ is a cyclic quadrilateral and $PR$ is a diameter. If $\angle RPQ={50}^{\circ }$ and $\angle SQP={45}^{\circ }$ then find $\angle QRS.$

• A. ${85}^{\circ }$
• B. ${90}^{\circ }$
• C. ${60}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

The correct option is A ${85}^{\circ }$

$\angle PQR={90}^o$ ....... (Angle subtended by diameter $={90}^o$)

$\angle RQS={90}^o-{45}^o={45}^o$

$\angle PRQ={180}^o-({50}^o+{90}^o)={40}^o$ ....... [sum of angles of triangle is ${180}^o$]

$\angle QRS={40}^o+{45}^o={85}^o$ .... [$\angle SRP={45}^o=\angle SQP$(Angle made on same base)]

Hence, ${85}^o$ is the correct answer.