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