In the figure,tangents PQ and PR are drawn to | vedclass

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(B) $PQ$ and $PR$ are two tangents drawn from an external point $P$.$	herefore PQ = PR$ [The lengths of tangents drawn from an external point to a ci

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(B) $PQ$ and $PR$ are two tangents drawn from an external point $P$.$	herefore PQ = PR$ [The lengths of tangents drawn from an external point to a circle are equal].In $	riangle PQR$,since $PQ = PR$,the angles opposite to equal sides are equal.$	herefore \angle PQR = \angle QRP$.In $	riangle PQR$,the sum of angles is $180^{\circ}$.$\angle PQR + \angle QRP + \angle RPQ = 180^{\circ}$$2 \angle PQR + 30^{\circ} = 180^{\circ}$$2 \angle PQR = 150^{\circ}$$\angle PQR = 75^{\circ}$.Since chord $SR \parallel$ tangent $PQ$,the alternate interior angles are equal.$	herefore \angle SRQ = \angle RQP = 75^{\circ}$.By the alternate segment theorem,the angle between the tangent $PR$ and chord $RQ$ is equal to the angle subtended by the chord in the alternate segment.$	herefore \angle PQR = \angle QSR = 75^{\circ}$.In $	riangle QRS$,the sum of angles is $180^{\circ}$.$\angle RQS + \angle QSR + \angle SRQ = 180^{\circ}$$\angle RQS + 75^{\circ} + 75^{\circ} = 180^{\circ}$$\angle RQS + 150^{\circ} = 180^{\circ}$$\angle RQS = 30^{\circ}$.

In the figure,tangents $PQ$ and $PR$ are drawn to a circle such that $\angle RPQ = 30^{\circ}$. $A$ chord $RS$ is drawn parallel to the tangent $PQ$. Find the $\angle RQS$. (in $^{\circ}$)

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