We are given a circle $PQRS$ with center $O$. We are also 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 are given a circle $PQRS$ with center $O$. We are also 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 know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Thus $\angle SPQ = \angle UQR = 68^\circ$.

Also, $\angle RPQ = 180^\circ - \angle TPS = 180^\circ - 74^\circ = 106^\circ$. This is because $\angle TPS$ is the exterior angle of cyclic quadrilateral $PQRS$ at vertex $P$.

$\angle SPQ = \angle SPR + \angle RPQ = 68^\circ$ is incorrect. It should be $\angle SPQ = \angle SRQ = 68^{\circ}$

The angle between the tangent and chord is equal to the angle in the alternate segment, so $\angle SPQ = \angle SRQ = 74^{\circ}$.

We also have that $\angle TPs = 74^{\circ}$, so $\angle SPQ = 180^{\circ} - 74^{\circ} = 106^{\circ}$.

Since the exterior angle equals the opposite interior angle of a cyclic quadrilateral, $\angle SRQ = 74^{\circ}$.

$\angle QSR = 40^{\circ}$

$\angle SRQ = \angle SRQ + \angle QSR = 74^{\circ}$

Also $\angle PQR = 180^{\circ} - \angle TPS = 180^{\circ} - 74^{\circ} = 106^{\circ}$.

Also, $\angle UQR = 68^{\circ}$, so $\angle PQS = 180^{\circ} - 68^{\circ} = 112^{\circ}$. But this is incorrect since $PQU$ is a straight line. Thus, $\angle PQR = 180^{\circ} - 68^{\circ} = 112^{\circ}$. Then $\angle PQS = 112^{\circ} - \angle RQS$. Also $\angle PRS = \angle PQS$.

Since $\angle UQR = 68^\circ$, the exterior angle theorem gives us that $\angle PQR + \angle UQR = 180^\circ$, hence $\angle PQR + 68^\circ = 180^\circ$ which means that $\angle PQR = 180^\circ - 68^\circ = 112^\circ$.

Also $\angle SPR = \angle SQR$. The angles subtended by the same arc.

$\angle PRQ = \angle PSQ$.

The sum of the angles in quadrilateral $PQRS$ is $360^{\circ}$.

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

$\angle SRQ = 74^\circ$. $\angle QSR = 40^{\circ}$.

In triangle $SQR$, $\angle SQR + \angle QRS + \angle RSQ = 180^\circ$.

$\angle QRS + 40^\circ = 74^\circ$. So, $\angle QRS = 34^{\circ}$.

$\angle PQR = \angle PQS + \angle SQR$. Thus, $\angle SQR = \angle PQR - \angle PQS = 112^{\circ} - \angle PQS$.

However, $\angle PRS = \angle PQS$ since they subtend the same arc $PS$.

Also, $\angle PQS = \angle PRS$ since they subtend the same arc $PS$.

In the quadrilateral $PQRS$, $\angle SPQ + \angle PQR + \angle QRS + \angle RSP = 360^\circ$

$106 + 112 + 74 + \angle RSP = 360^\circ$.

$292 + \angle RSP = 360^\circ$.

$\angle RSP = 360 - 292 = 68^{\circ}$

Since $\angle QSR = 40^\circ$, then $\angle PSR = \angle QSR + \angle PSQ = 40 + \angle PSQ = 68^\circ$.

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

Since $\angle PSQ = \angle PRQ$, then $\angle PRQ = 28^\circ$.

Now we have $\angle QRS = \angle QRP + \angle PRS = 34^\circ$, and $\angle PRQ = 28^\circ$, then

$28^\circ + \angle PRS = 34^\circ$.

So $\angle PRS = 34 - 28 = 6^\circ$.

3. Final Answer

The value of $\angle PRS$ is $6^\circ$.