Question

In Fig. 8.66, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that $\angle RPQ={30}^{\circ }$. A chord RS is drawn parallel to the tangent PQ. Find $\angle RQS$.

Answer 1

Given $\angle RPQ=30°$ and PR and PQ are tangents drawn from P to the same circle.
Hence PR = PQ [Since tangents drawn from an external point to a circle are equal in length]
$\therefore$ $\angle PRQ=\angle PQR$ [Angles opposite to equal sides are equal in a triangle]

In ΔPQR

$\angle RQP+\angle QRP+\angle RPQ=180°$[Angle sum property of a triangle]
$2\angle RQP+30°=180°$

$2\angle RQP=150°$

$\angle RQP=75°$

Hence, $\angle RQP=\angle QRP=75°$ and $\angle RQP=\angle RSQ=75°$ [ By Alternate Segment Theorem]
Given, $RS||PQ$ Therefore $\angle RQP=\angle SRQ=75°$ [Alternate angles]
$\angle RSQ=\angle SRQ=75°$

$\therefore$$QRS$ is also an isosceles triangle. [Since sides opposite to equal angles of a triangle are equal.]

$\angle RSQ+\angle SRQ+\angle RQS=180°$[Angle sum property of a triangle]

$75°+75°+\angle RQS=180°$
$150°+\angle RQS=180°$
$\therefore$ $\angle RQS=30°$