In circle $PQRS$ with center $O$, $\angle PQR = 72^\circ$ and $OR \parallel PS$. We need to find the measure of $\angle OPS$.

1. Problem Description

In circle $PQRS$ with center $O$, $\angle PQR = 72^\circ$ and $OR \parallel PS$. We need to find the measure of $\angle OPS$.

2. Solution Steps

First, the angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any point on the remaining part of the circle.

$\angle POR = 2 \angle PQR$

$\angle POR = 2(72^\circ) = 144^\circ$

Since $OR \parallel PS$, $\angle OPS = \angle POR$, because they are alternate interior angles. However, this assumes $OR$ and $PS$ are cut by a transversal, which they are not here. Instead, we can consider supplementary angles.

Let $\angle OPS = x$. Since $OR \parallel PS$, $\angle ROS = \angle ORS$.

Also, $\angle ROS + \angle OPS + \angle PSO = 180$, this would work if $ROS$ was a straight line, which it is not.

Alternatively, consider that $OR \parallel PS$, so the consecutive interior angles $\angle SOP$ and $\angle ROS$ have the property that the angles sum to $180^\circ$. So, the sum of the interior angles on one side of the transversal is $180^\circ$.

Since $OR \parallel PS$, then $\angle PSO = \angle ROS$ (alternate interior angles). We have $OP = OS$ as they are radii of the circle, so $\triangle OPS$ is isosceles, and $\angle OPS = \angle OSP$. Let $\angle OPS = x$, then $\angle OSP = x$.

Now, $\angle POS + \angle OPS + \angle OSP = 180^\circ$

$\angle POS + x + x = 180^\circ$

$\angle POS = 180^\circ - 2x$

Since $OR \parallel PS$, $\angle ROS = \angle PSO = \angle OPS = x$.

We know that $\angle POR = 144^\circ$.

Also $\angle POS + \angle ROS = \angle POR$, however this should be $\angle POS + \angle ROS = \angle PSR$ where PSR are colinear which is not true here.

Consider quadrilateral $ORSP$. The sum of angles in a quadrilateral is $360^\circ$. Since $OR \parallel PS$, $\angle ORS + \angle RSP = 180^\circ$. We have $\angle OPS = \angle OSP = x$. Also $\angle RSP = \angle RSO + \angle OSP = \angle RSO + x$ and $\angle ORS = \angle ORO$

We have $\angle POS = 180-2x$ and $\angle POR = 144$.

Then $\angle ROS = \angle POR - \angle POS$.

So we should instead use the sum of the arcs =

3

6

0. arc$PR = 2*72 = 144$, arc$QR =144$

arc$RS= arc RS$, arc$SP=2*OPS$

arc RP = 2 *72 =144

arcs $OR // SP$ and $\angle ROS = 36$. and $OPS=36$.

3. Final Answer

36°