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°


Answer:

^@ 60^\circ ^@

Step by Step Explanation:
  1. Let us join ^@ OQ ^@ and ^@ OR ^@. Also, produce ^@ PQ ^@ and ^@ PR ^@ to ^@ M ^@ and ^@ N ^@ respectively.
    R O N M P Q S 60°
  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 ^@.

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