The vertices of a $$\triangle$$PQR lie on a circle with centre O. SR is a tangent to the circle at the point R. If QR bisects the $$\angle$$ORS, then what is the measure of $$\angle$$RPQ?

As per the given in the question,

It is given that, SR is the tangent on the circle and RQ is the angle bisector of the circle.

So,$$ 2x=90$$

Hence,$$ x=45^\circ$$

In  $$\triangle$$ORQ,

OR=OQ    (radius of the circle)

Hence,

$$\Rightarrow \angle ORQ=\angle OQR=45^\circ$$

now,

$$\Rightarrow \angle ORQ+\angle OQR+\angle ROQ=180$$

$$\Rightarrow 45+45+ \angle ROQ=180$$

$$\Rightarrow \angle ROQ=180-90=90^\circ$$

We know that, in any circle, angle subtended by the minor arc on center is twice angle subtended by on the circumference.

So

$$\Rightarrow \angle ROQ=2 \angle RPQ$$

$$\Rightarrow \angle RPQ=\dfrac{90}{2}=45^\circ$$