Question

In Fig. 7, 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

We know that tangents from an external point are equal in length.
$\therefore$ PQ = PR
In $\Delta$ PQR,
PQ = PR
$\therefore \angle PQR=\angle PRQ$ (Angles opposite to equal sides are equal).
Now in $\Delta$ PQR,
$\angle PQR+\angle PRQ+\angle RQP={180}^{\circ }2\angle RQP={180}^{\circ }-{30}^{\circ }\angle RQP={75}^{\circ }$
Also, radius is perpendicular to the tangent at the point of contact.
$\therefore \angle OQP=\angle ORP={90}^{\circ }$
Now, in PQOR,
$\angle ROQ+\angle OQP+\angle QPR+\angle PRO={360}^{\circ }{90}^{\circ }+{90}^{\circ }+{30}^{\circ }+\angle ROQ={360}^{\circ }\angle ROQ={150}^{\circ }$
Since, angle subtended by an arc at any point on the circle is half the angle subtended at the centre by the same arc.
angle QSR = ${75}^{\circ }$
Also, $\angle QSR=\angle SQT$ (Alternate interior angles)
$\therefore \angle SQT={75}^{\circ }$
Now,
$\angle SQT+\angle PQR+\angle SQR={180}^{\circ }{75}^{\circ }+{75}^{\circ }\angle SQR={180}^{\circ }\angle SQR={30}^{\circ }$