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 ^@.
Answer:
^@ 60^\circ ^@
- Let us join ^@ OQ ^@ and ^@ OR ^@. Also, produce ^@ PQ ^@ and ^@ PR ^@ to ^@ M ^@ and ^@ N ^@ respectively.
- 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} @^ - 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] } @^ - 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 ^@ - Thus, the measure of ^@ \angle RQS ^@ is ^@ 60^\circ ^@.
