### In the given figure, tangents $PQ$ and $PR$ are drawn from an external point $P$ to a circle with center $O$, such that $\angle RPQ = 60^\circ$. A chord $RS$ is drawn parallel to the tangent $PQ$. Find the measure of $\angle RQS$. R O P Q S 60°

$60^\circ$

Step by Step Explanation:
1. Let us join $OQ$ and $OR$. Also, produce $PQ$ and $PR$ to $M$ and $N$ respectively.
2. We know that the angle between two tangents from an external point is supplementary to the angle subtended by the radii at the center.
Thus, \begin{aligned} & \angle RPQ + \angle ROQ = 180^\circ \implies \angle ROQ = 180^\circ - \angle RPQ = 180^\circ - 60^\circ = 120^\circ \end{aligned}
3. We also know that the angle subtended by an arc at the center is twice the angle subtended by the same arc on the remaining part of the circle.
So, $$\angle RSQ = \dfrac { 1 } { 2 } \angle ROQ = \dfrac { 1 } { 2 } \times 120^\circ = 60^\circ$$ As, $RS//PQ,$ $$\implies \angle SQM = \angle RSQ = 60^\circ \text{ [Alternate Interior Angles] }$$ Also, $$\angle PQR = \angle RSQ = 60^\circ \text{ [Alternate Segment Theorem] }$$
4. We know that the sum of angles on a straight line is $180^\circ$.

As $PM$ is a straight line. $$\implies \angle SQM + \angle RQS + \angle PQR = 180^\circ$$ Therefore, $\angle RQS = 180^\circ - (\angle SQM + \angle PQR) = 180^\circ - (60^\circ + 60^\circ) = 60^\circ$
5. Thus, the measure of $\angle RQS$ is $60^\circ$. 