Given,

$\angle TRQ={{30}^{\circ }}$ .

At point R, OR ⊥ RQ.

So, $\angle ORQ={{90}^{\circ }}$

$\Rightarrow \angle TRQ+\angle ORT={{90}^{\circ }}$

$\Rightarrow \angle ORT={{90}^{\circ }}-{{30}^{\circ }}={{60}^{\circ }}$

It’s seen that, ST is the diameter,

So, $\angle SRT={{90}^{\circ }}$  [ ∵ Angle in semicircle = 90°]

Then,

$\angle ORT+\angle SRO={{90}^{\circ }}$

$\angle SRO+\angle PRS={{90}^{\circ }}$

$\therefore \angle PRS={{90}^{\circ }}-{{30}^{\circ }}={{60}^{\circ }}$