Question

In the given figure, $POQ$ is the diameter of the circle with center $O$. Quadrilateral $PQRS$ is a cyclic quadrilateral and $SQ$ is joined. If $\angle R={138}^{\circ }$, then $m\angle PQS$ is:

• A. ${90}^{\circ }$
• B. ${42}^{\circ }$
• C. ${48}^{\circ }$
• D. ${38}^{\circ }$

Answer 1

The correct option is C ${48}^{\circ }$

Given- $\overline{POQ}$ is a diameter of a given circle. $PQRS$ is a cyclic quadrilateral. $SQ$ is joined. Also, $\angle SRQ={138}^{\circ }$.

Since, $\overline{POQ}$ is the diameter of the given circle, it subtends $\angle PSQ$ to the circumference at $S$, i.e. $\angle PSQ={90}^{\circ }$ ...[ since it is an angle in a semicircle].

Again,

$\angle QPS+\angle QRS={180}^{\circ }$ ...[ sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$]

$\Rightarrow \angle QPS={180}^{\circ }-\angle QRS$

$\Rightarrow \angle QPS={180}^{\circ }-{138}^{\circ }$

$\Rightarrow \angle QPS={42}^{\circ }$.

In $△PSQ,$

$\angle PQS+\angle PSQ+\angle QPS={180}^{\circ }$

$\angle PQS={180}^{\circ }-(\angle PSQ+\angle QPS)$

$={180}^{\circ }-({90}^{\circ }+{42}^{\circ })$

$={48}^{\circ }$

Hence, option $C$ is correct.