In the figure,if OP parallel RS, angle OPQ = | vedclass

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(D) To solve this,draw a line $XY$ parallel to $OP$ passing through point $R.$ Since $OP \parallel RS,$ it follows that $XY \parallel OP \parallel RS.

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$60$

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(D) To solve this,draw a line $XY$ parallel to $OP$ passing through point $R.$ Since $OP \parallel RS,$ it follows that $XY \parallel OP \parallel RS.$Now,consider the parallel lines $OP$ and $XY.$ Since $PQ$ is a transversal,the sum of consecutive interior angles is $180^{\circ}.$$\angle OPQ + \angle PQR_{1} = 180^{\circ}$$110^{\circ} + \angle PQR_{1} = 180^{\circ}$$\angle PQR_{1} = 180^{\circ} - 110^{\circ} = 70^{\circ}$Next,consider the parallel lines $XY$ and $RS.$ Since $QR$ is a transversal,the sum of consecutive interior angles is $180^{\circ}.$$\angle QRS + \angle RQR_{2} = 180^{\circ}$$130^{\circ} + \angle RQR_{2} = 180^{\circ}$$\angle RQR_{2} = 180^{\circ} - 130^{\circ} = 50^{\circ}$Finally,$\angle PQR = \angle PQR_{1} + \angle RQR_{2} = 70^{\circ} + 50^{\circ} = 120^{\circ}.$Wait,re-evaluating the geometry: If we extend $OP$ to intersect $QR$ at $X,$ then $\angle RXP = 180^{\circ} - 130^{\circ} = 50^{\circ}$ (interior angles on same side of transversal). In $	riangle PXQ,$ $\angle OPQ$ is an exterior angle,so $\angle OPQ = \angle PXQ + \angle PQR.$$110^{\circ} = 50^{\circ} + \angle PQR$$\angle PQR = 110^{\circ} - 50^{\circ} = 60^{\circ}.$

In the figure,if $OP \parallel RS,$ $\angle OPQ = 110^{\circ}$ and $\angle QRS = 130^{\circ},$ then $\angle PQR$ is equal to: (in $^{\circ}$)

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