Question

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

• A. ${85}^{\circ }$
• B. ${135}^{\circ }$
• C. ${145}^{\circ }$
• D. ${110}^{\circ }$

Answer 1

The correct option is C ${145}^{\circ }$

We have PQ || RS. Produce PQ to M.
$\angle CQP=\angle MQD$ [Vertically opp. $\angle S$]
$\therefore {60}^{\circ }=\angle 1+{25}^{\circ }$
$\Rightarrow \angle 1={35}^{\circ }$
Now, QM || RS and QR cuts CD and AB
$\angle ARQ=\angle RQD={25}^{\circ }$ [Alt. $\angle S$]
$\therefore \angle 1+(\angle ARQ+\angle RMP)={180}^{\circ }$
$\Rightarrow \angle 1+(\angle ARQ+\angle ARS)={180}^{\circ }$
$\Rightarrow {35}^{\circ }+({25}^{\circ }+\angle ARS)={180}^{\circ }$
$\Rightarrow \angle ARS={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\therefore \angle QRS=\angle ARQ+\angle ARS$
$={25}^{\circ }+{120}^{\circ }={145}^{\circ }$