Question

$PQ$ is a diameter and $PQRS$ is a cyclic quadrilateral. If $\angle PSR={150}^o$, then measure of $\angle RPQ$ is:

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${30}^{\circ }$
• D. None of these

Answer 1

The correct option is B ${60}^{\circ }$

$PQRS$ is a cyclic quadrilateral.

Then, $\angle PQR+\angle PSR={180}^{\circ }$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$⟹$ $\angle PQR+{150}^o={180}^{\circ }$

$⟹$ $\angle PQR={180}^{\circ }-{150}^{\circ }={30}^{\circ }$.

In $△PQR$,
$\angle PRQ={90}^{\circ }$ (Angle of a semicircle)
Then, $\angle RPQ+{90}^{\circ }+{30}^{\circ }={180}^{\circ }$ ...[Angle sum property]

$\Rightarrow \angle RPQ+{120}^{\circ }={180}^{\circ }$
$\Rightarrow \angle RPQ={60}^{\circ }$.

Hence, option $B$ is correct.