The problem provides a circle $PQRS$ with center $O$. We are given that $\angle UQR = 68^\circ$, $\angle TPS = 74^\circ$, and $\angle QSR = 40^\circ$. The goal is to find the value of $\angle PRS$.

1. Problem Description

The problem provides a circle $PQRS$ with center $O$. We are given that $\angle UQR = 68^\circ$, $\angle TPS = 74^\circ$, and $\angle QSR = 40^\circ$. The goal is to find the value of $\angle PRS$.

2. Solution Steps

First, we use the property that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Since $PQRS$ is a cyclic quadrilateral and $UQR$ is an exterior angle at $Q$, we have $\angle UPS = \angle UQR = 68^\circ$.

Then, since $\angle TPS = 74^\circ$, we can find $\angle SPQ$ by noting that $\angle TPS + \angle SPQ = 180^\circ$ (straight line). This is not correct. Since $TPS$ is a straight line, $\angle TPS + \angle SPQ = 180^\circ$ does not hold true here. We have $\angle TPS$ and $\angle UPS$ and we can use this to find $\angle TPU$

Since $\angle TPS = 74^\circ$ and $\angle UPS = 68^\circ$, we use the fact that $\angle SPQ$ is supplementary to $\angle TPS$, where $T,P,U$ are collinear.

$\angle SPQ = 180^\circ - 74^\circ = 106^\circ$.

Since $PQRS$ is a cyclic quadrilateral, $\angle SPQ + \angle SRQ = 180^\circ$ (opposite angles of a cyclic quadrilateral are supplementary).

Therefore, $\angle SRQ = 180^\circ - \angle SPQ = 180^\circ - 106^\circ = 74^\circ$.

We are given $\angle QSR = 40^\circ$.

Now, $\angle SRQ = \angle QSR + \angle PRS$, so $\angle PRS = \angle SRQ - \angle QSR$.

$\angle PRS = 74^\circ - 40^\circ = 34^\circ$.

3. Final Answer

34