To solve the problem:

Given:

1. $O$ is the center of the circle.
2. $PQ = QR = RS$ (i.e., the sides are equal, forming a cyclic quadrilateral).
3. $\angle PTS = 75^\circ$.

We need to find $\angle PQR$.

Step 1: Recognize the geometry

Since $PQ = QR = RS$, the quadrilateral $PQRS$ is a cyclic quadrilateral with equal chords. Therefore, all the angles subtended by equal chords will have equal measures.

Step 2: Use the cyclic quadrilateral property

In a cyclic quadrilateral: $\text{Opposite angles are supplementary.}$ For $PQRS$, $\angle PTS + \angle PQR = 180^\circ$.

Given $\angle PTS = 75^\circ$, $\angle PQR = 180^\circ - 75^\circ = 105^\circ.$

Final Answer:

$\boxed{\angle PQR = 105^\circ.}$

Let me know if you need further clarifications or detailed explanations! Here are 5 related questions:

1. What are the properties of cyclic quadrilaterals in a circle?
2. How do equal chords subtend equal angles at the center or circumference?
3. Can the result change if $PQ \neq QR \neq RS$?
4. How does the concept of supplementary angles apply to cyclic quadrilaterals?
5. What is the difference between inscribed and central angles subtended by the same arc?

Tip: Always identify symmetry and equal segments in circle problems for quick insights!