Question

In figure, $PQRS$ is a rectangle. If $\angle RPQ={30}^{\circ }$ then the value of $(x+y)$ is

• A. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${150}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is D ${180}^{\circ }$

Let the point of intersection of the diagonals of the rectangle be O.

Given $\angle OPQ=30°$

$\therefore \angle SPO=60°$ Similarly $\angle RQO=60°$

As its a rectangle, all the angles are $90°.$

$\therefore \angle PQR=\angle PQO+\angle OQR$

$=30°+60°$

$=90°$

$\therefore \angle OQR=60°$ i.e, $x=60°$

From $△POQ$

Sum of angles of $△$les $=180°$

$\Rightarrow \angle OPQ+\angle PQO+\angle POQ=180°$

$\therefore \angle POQ=180°-{30}^o-{30}^o=120°$

i.e, $\angle POQ=\angle SOR=120°=y$ [ i.e, opposite angles ]

$\therefore$ Sum of $x+y=120°+60°=180°.$

Hence, the answer is $180°.$