Question

In the given figure, if AB || CD || EF, PQ || RS, $\angle RQD={25}^{\circ }$ and $\angle CQP={60}^{\circ }$, then $\angle ARS$ is equal to

• A. ${120}^{\circ }$
• B. ${110}^{\circ }$
• C. ${85}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is A ${120}^{\circ }$

We have PQ || RS. Produce PQ to M.
$\angle CQP=\angle MQD$ [Vertically opp. $\angle S$]
$\therefore {60}^{\circ }=\angle 1+{25}^{\circ }$
$\Rightarrow \angle 1={35}^{\circ }$

We have PQ || RS. Produce PQ to M.

$\angle CQP=\angle MQD$ [Vertically opp. angles]
$\therefore {60}^{\circ }=\angle 1+{25}^{\circ }$
$\Rightarrow \angle 1={35}^{\circ }$

$\angle AMQ=\angle 1+\angle RQD={60}^{\circ }$ [alternate angles]

$\angle RMQ+\angle AMQ={180}^{\circ }$
$\Rightarrow \angle RMQ=180-60={120}^{\circ }$

Since PQ || RS,
$\angle RMQ=\angle ARS={120}^{\circ }$ [alternate angles]