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

1. Problem Description

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

2. Solution Steps

First, we find $\angle PQR$ using the exterior angle property. Since $\angle UQR = 68^\circ$, then $\angle PQR = 180^\circ - 68^\circ = 112^\circ$.

Next, we find $\angle RSP$ using the exterior angle property. Since $\angle TPS = 74^\circ$, then $\angle RSP = 180^\circ - 74^\circ = 106^\circ$.

Since $PQRS$ is a cyclic quadrilateral, opposite angles are supplementary. Thus, $\angle PQR + \angle PSR = 180^\circ$ and $\angle QPS + \angle QRS = 180^\circ$.

Also, $\angle PSR = \angle QSR + \angle PSQ$, so $\angle PSQ = \angle PSR - \angle QSR = 106^\circ - 40^\circ = 66^\circ$.

Now, consider quadrilateral $PQRS$. Since it is a cyclic quadrilateral, $\angle QRS + \angle QPS = 180^\circ$.

Also, $\angle QRS = \angle QRP + \angle PRS$.

Since $\angle TPS = 74^\circ$, $\angle SRQ = 180^\circ - \angle SPQ$. Since $PQRS$ is cyclic, $\angle SPQ$ and $\angle SRQ$ sum to $180^\circ$. Also since $TPS$ form a line we have $\angle SPQ+\angle TPS=180^\circ$, $\angle SPQ = 180^\circ-74^\circ = 106^\circ$.

This implies that $\angle SRQ = 180^\circ-106^\circ=74^\circ$.

We also know that $\angle QSR = 40^\circ$, so $\angle PRS = \angle SRQ-\angle QSR = 74^\circ-40^\circ=34^\circ$.

However, using $\angle PQR=112^\circ$ and $\angle PSR = 106^\circ$ we have $\angle PQR + \angle PSR = 112 + 106 = 218^\circ \neq 180^\circ$.

Therefore, we proceed with the following. Since $\angle TPS = 74^\circ$, $\angle SPQ=180-74=106^\circ$. Since $\angle PQR + \angle PSR =180$ and $\angle QSR=40^\circ$, we also have $\angle PSQ = \angle PSR -40$. Also, $\angle PQR + \angle PSR=180$. Also, $\angle PQR = 180-68 = 112^\circ$. Since $\angle UQR = 68^\circ$, $\angle PQR = 180^\circ - 68^\circ = 112^\circ$.

Since $\angle QRS + \angle QPS = 180^\circ$, we need to calculate $\angle QPS$. Also, $\angle PSR = 180 - \angle PQR = 180 - 112 = 68^\circ$.

Then $\angle PSQ = \angle PSR - \angle QSR = 68 - 40 = 28^\circ$.

In triangle $PSQ$, $\angle PQS = 180 - \angle SPQ - \angle PSQ = 180 - 106 - 28 = 46^\circ$.

Since $\angle PQR = 112^\circ$, $\angle PQR = \angle PQS + \angle SQR = 112^\circ$.

Therefore, $\angle SQR = 112 - 46 = 66^\circ$.

The angles in triangle $SQR$ are $\angle SQR=66, \angle QSR=40$, therefore $\angle QRS = 180-66-40 = 74$. Thus $\angle PRS = 74-x$

We have $\angle QPS = 106^\circ$ and $\angle QRS = 180^\circ - 106^\circ = 74^\circ$.

Thus, $\angle PRS = \angle QRS - \angle QRP$, but since $\angle QSR = 40^\circ$ it cannot exceed $74^\circ$.

Therefore the solution given earlier of $\angle SRQ=180^\circ-\angle SPQ = 74^\circ$.

Then since $\angle QSR=40^\circ$, $\angle PRS=34^\circ$.

3. Final Answer

34