PQRS is a cyclic quadrilateral. TPX is a tangent. RQX is a straight line. RPQ = 50^ . PRS = 25^ . If PQ = RS, find PRQ.

Angles subtended by the same arc at the circumference are equal.
$\angle PRQ$ and $\angle PSQ$ are subtended by arc PQ.
$\angle RPQ$ and $\angle RSQ$ are subtended by arc RQ.
$\angle SPR$ and $\angle SQR$ are subtended by arc SR.
$\angle QPS$ and $\angle QRS$ are subtended by arc QS.

We are given $\angle RPQ = 50^\circ$.
$\angle PRQ$ and $\angle PSQ$ are subtended by arc PQ.
$\angle SPR$ and $\angle SQR$ are subtended by arc SR.
We are given $\angle PRS = 25^\circ$.

Consider $\triangle PQR$. The sum of angles in a triangle is $180^\circ$.
We need to find $\angle PRQ$. We know $\angle RPQ = 50^\circ$.
We also know that $\angle QPS = \angle QRS$ (angles subtended by arc QS).
And $\angle PQR + \angle PSR = 180^\circ$ (opposite angles of a cyclic quadrilateral).

Let's re-examine the diagram and given values.
The diagram shows $\angle QPR = 50^\circ$ and $\angle PRS = 25^\circ$.
The markings on the chords PQ and RS indicate that PQ = RS.
If PQ = RS, then the angles subtended by these chords at the circumference are equal.
So, $\angle PRQ = \angle SPR$.
Also, $\angle PSQ = \angle QPR = 50^\circ$ (angles subtended by arc PQ).
And $\angle RPS = \angle RQS$ (angles subtended by arc RS).

Since PQ = RS, the arcs PQ and RS are equal.
Therefore, the angles subtended by these arcs at the circumference are equal.
$\angle PRQ = \angle SPR$.
We are given $\angle RPQ = 50^\circ$ and $\angle PRS = 25^\circ$.
In $\triangle PRS$, we have $\angle PRS = 25^\circ$.
In $\triangle PQR$, we have $\angle RPQ = 50^\circ$.

Let's use the property that angles subtended by the same arc are equal.
$\angle PRQ$ is subtended by arc PQ.
$\angle PSQ$ is subtended by arc PQ.
So, $\angle PRQ = \angle PSQ$.

We are given $\angle RPQ = 50^\circ$.
We are given $\angle PRS = 25^\circ$.
From the diagram, the chords PQ and RS are marked as equal.
If chords PQ and RS are equal, then the angles they subtend at the circumference are equal.
So, $\angle PRQ = \angle SPR$.
Also, $\angle PSQ = \angle QPR = 50^\circ$.
And $\angle RPS = \angle RQS$.

Since PQ = RS, then $\angle PRQ = \angle SPR$.
In $\triangle PQR$, we have $\angle RPQ = 50^\circ$.
In $\triangle PRS$, we have $\angle PRS = 25^\circ$.

Let's use the fact that $\angle PSQ = \angle PRQ$ (angles subtended by the same arc PQ).
Also, $\angle RPS = \angle RQS$ (angles subtended by the same arc RS).
Since PQ = RS, then $\angle PRQ = \angle RPS$.
So, $\angle PRQ = \angle RPS$.

We know $\angle QPS = \angle QPR + \angle RPS = 50^\circ + \angle RPS$.
We also know $\angle QRS = \angle QRP + \angle PRS = \angle QRP + 25^\circ$.
Since PQRS is a cyclic quadrilateral, $\angle QPS = \angle QRS$.
So, $50^\circ + \angle RPS = \angle QRP + 25^\circ$.
$25^\circ + \angle RPS = \angle QRP$.

Since PQ = RS, then $\angle PRQ = \angle SPR$.
Let $\angle PRQ = x$. Then $\angle SPR = x$.
In $\triangle PQR$, $\angle PQR + \angle QRP + \angle RPQ = 180^\circ$.
$\angle PQR + x + 50^\circ = 180^\circ$.
$\angle PQR = 130^\circ - x$.

In $\triangle PRS$, $\angle PSR + \angle SPR + \angle PRS = 180^\circ$.
$\angle PSR + x + 25^\circ = 180^\circ$.
$\angle PSR = 155^\circ - x$.

Since PQRS is a cyclic quadrilateral, $\angle PQR + \angle PSR = 180^\circ$.
$(130^\circ - x) + (155^\circ - x) = 180^\circ$.
$285^\circ - 2x = 180^\circ$.
$2x = 285^\circ - 180^\circ$.
$2x = 105^\circ$.
$x = 52.5^\circ$.

So, $\angle PRQ = 52.5^\circ$.
Reason: Angles subtended by equal chords (PQ and RS) are equal, and opposite angles of a cyclic quadrilateral sum to $180^\circ$.

From part (a), we found $x = \angle PRQ = 52.5^\circ$.
We also established that $\angle PSR = 155^\circ - x$.
Substitute $x = 52.5^\circ$:
$\angle PSR = 155^\circ - 52.5^\circ = 102.5^\circ$.
Reason: Opposite angles of a cyclic quadrilateral sum to $180^\circ$.
Alternatively, $\angle PSR = \angle PQR + \angle QRS$.
We know $\angle PQR = 130^\circ - x = 130^\circ - 52.5^\circ = 77.5^\circ$.
$\angle PSR = 180^\circ - \angle PQR = 180^\circ - 77.5^\circ = 102.5^\circ$.

Consider $\triangle PQX$. We need to find $\angle PXQ$.
We know $\angle RPQ = 50^\circ$.
We know RQX is a straight line.
$\angle PQR = 77.5^\circ$ (from part b calculation).
So, $\angle PQX = 180^\circ - \angle PQR = 180^\circ - 77.5^\circ = 102.5^\circ$ (angles on a straight line).

In $\triangle PQX$, $\angle PXQ + \angle QPX + \angle PQX = 180^\circ$.
We need $\angle QPX$.
TPX is a tangent at P.
$\angle QPX$ is the angle between the tangent PX and the chord PQ.
By the Alternate Segment Theorem, $\angle QPX = \angle PRQ$.
From part (a), $\angle PRQ = 52.5^\circ$.
So, $\angle QPX = 52.5^\circ$.

Now substitute the known angles into the sum of angles for $\triangle PQX$:
$\angle PXQ + 52.5^\circ + 102.5^\circ = 180^\circ$.
$\angle PXQ + 155^\circ = 180^\circ$.
$\angle PXQ = 180^\circ - 155^\circ = 25^\circ$.
Reason: Alternate Segment Theorem and sum of angles in a triangle.

$\angle TPS$ is the angle between the tangent TP and the chord PS.
By the Alternate Segment Theorem, $\angle TPS = \angle PRS$.
We are given $\angle PRS = 25^\circ$.
Reason: Alternate Segment Theorem.

$\angle POS$ is the angle subtended by arc PS at the centre O.
$\angle PRS$ is the angle subtended by arc PS at the circumference.
The angle at the centre is twice the angle at the circumference subtended by the same arc.
So, $\angle POS = 2 \times \angle PRS$.
We are given $\angle PRS = 25^\circ$.
$\angle POS = 2 \times 25^\circ = 50^\circ$.
Reason: Angle at the centre is twice the angle at the circumference.