Question

$C$ is the centre of a circle passing through the points $P,Q,R$and$S$ taken in order. If $\angle PSR={120}^{\circ }$ and $PQ$ is a diameter, then $\angle QPR=$

• A. ${30}^{\circ }$
• B. ${40}^{\circ }$
• C. ${45}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

Refer figure:

Given that: $\angle PSR=120°$

$\angle PRQ=90°$ (Diameter of the circle subtends an angle of 90°)

Consider cyclic quadrilateral $PQRS$:

$\angle PSR+\angle PQR=180°$ (sum of opposite angles of a cyclic quadrilateral is 180°)

Thus, $\angle PQR=180°-120°=60°$

In $△PQR$:

$\angle QPR=180°-\angle PRQ-\angle PQR$ .....(Sum of three angles of a triangle is 180°)

$\angle QPR=180°-90°-60°=30°$

Answer: $\angle QPR=30°$