Question

In Fig. 10.70, 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

Since tangents drawn from an external point to a circle are equal.

So, $PQ=PR$

Also, $\angle RQP=\angle QRP$ [Angle opposite to equal sides]
$\angle RQP+\angle QRP+\angle RPQ={180}^{\circ }$ [Angle sum property of a triangle]
$2\angle RQP+{30}^{\circ }={180}^{\circ }$
$2\angle RQP={150}^{\circ }$
$\angle RQP=\angle QRP={75}^{\circ }$
$\angle RQP=\angle RSQ={75}^{\circ }$ [ Angles in alternate Segment Theorem states that angle between chord and tangent is equal to the angle in the alternate segment]
RS is parallel to PQ
Therefore $\angle RQP=\angle SRQ={75}^{\circ }$ [Alternate angles]
$\angle RSQ=\angle SRQ={75}^{\circ }$
Now, in triangle QRS
$\angle RSQ+\angle SRQ+\angle RQS={180}^{\circ }$ [Angle sum property of a triangle]
${75}^{\circ }+{75}^{\circ }+\angle RQS={180}^{\circ }$
$150°+\angle RQS={180}^{\circ }$
$\angle RQS={30}^{\circ }$.