From the given diagram and the problem description:

Key Information:

1. QS = RS: This makes triangle $\triangle QRS$ an isosceles triangle.
2. $\angle SQR = 35^\circ$.
3. The point $O$ is the center of the circle.

We are tasked to find:

• (a) $\angle QSR$,
• (b) $\angle QPR$,
• (c) $\angle TQR$.

Step-by-Step Solution:

(a) $\angle QSR$

Since $\triangle QRS$ is isosceles ($QS = RS$):

• The base angles of the isosceles triangle are equal: $\angle SQR = \angle QSR.$
• Given $\angle SQR = 35^\circ$, therefore: $\angle QSR = 35^\circ.$

(b) $\angle QPR$

The total angle of $\triangle QRS$ is: $\angle QRS + \angle SQR + \angle QSR = 180^\circ.$ Substitute the known values: $\angle QRS + 35^\circ + 35^\circ = 180^\circ.$ $\angle QRS = 110^\circ.$ Now, since $O$ is the center of the circle, $\angle QPR$ (an angle at the circumference) subtends the same arc as $\angle QRS$ (the central angle). By the angle at center theorem: $\angle QPR = \frac{1}{2} \angle QRS.$ Substitute $\angle QRS = 110^\circ$: $\angle QPR = \frac{1}{2} \times 110^\circ = 55^\circ.$

(c) $\angle TQR$

The quadrilateral $QTRP$ is a cyclic quadrilateral because all its vertices lie on the circle. In a cyclic quadrilateral, opposite angles sum to $180^\circ$. Thus: $\angle QPR + \angle TQR = 180^\circ.$ Substitute $\angle QPR = 55^\circ$: $55^\circ + \angle TQR = 180^\circ.$ $\angle TQR = 180^\circ - 55^\circ = 125^\circ.$

Final Answers:

(a) $\angle QSR = 35^\circ$.
(b) $\angle QPR = 55^\circ$.
(c) $\angle TQR = 125^\circ$.

Would you like further explanation of any step?

Here are 5 related questions to explore:

1. How do we prove the angle at the center theorem?
2. What are the properties of cyclic quadrilaterals?
3. Why do isosceles triangles have equal base angles?
4. How can we generalize the relation between central and inscribed angles in a circle?
5. What is the significance of opposite angles in cyclic quadrilaterals?

Tip: Always look for symmetry and cyclic properties when dealing with geometry in circles!