Question

Question 7
In the figure, $OP||RS,\angle OPQ={110}^{\circ }and\angle QRS={130}^{\circ },then\angle PQR$ is equal to:


A) ${40}^{\circ }$
B) ${50}^{\circ }$
C) ${60}^{\circ }$
​​​​​​​D) ${70}^{\circ }$

Answer 1

The answer is C.
In the given figure, producing OP, to intersect RQ at X. Since, OP ∥RS and RX is a transversal.
So, $\angle RXP=\angle XRS$ [alternate angles]

$\Rightarrow \angle RXP={130}^{\circ }[\because \angle QRS={130}^{\circ }(given\left)\right].....\left(i\right)$
Now, RQ is all line segment.
$So,\angle PXQ+\angle RXP={180}^{\circ }$ [line pair axiom]
$\Rightarrow \angle PXQ={180}^{\circ }-\angle RXP={180}^{\circ }-{130}^{\circ }$ [From Eq. (i)]
$\Rightarrow \angle PXQ={50}^{\circ }$
In $\Delta PQX,\angle OPQ$ is an exterior angle,
$\therefore \angle OPQ=\angle PXQ+\angle PQX$
[$\because$ exterior angle = sum of two opposite interior angles]
$\Rightarrow {110}^{\circ }={50}^{\circ }+\angle PQX$
$\Rightarrow \angle PQX={110}^{\circ }-{50}^{\circ }$
$\therefore \angle PQR={60}^{\circ }[\because \angle PQX=\angle PQR]$