question_answer

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:

We have,           \[PR=PQ\]

and              \[\angle PRQ=\angle PQR\]

In \[\Delta \text{ }PQR\],

\[\angle PRQ+\angle PQR+\angle RPQ=180{}^\circ \]

\[2\angle PRQ+30{}^\circ =180{}^\circ \]

\[\angle PRQ=\frac{180{}^\circ -30{}^\circ }{2}=75{}^\circ \]

\[\because \,\,~SR\parallel QP\] and QR is a transversal

\[\angle SRQ=\angle PQR=75{}^\circ \]

Join \[OR,OQ\].

\[\therefore \angle ORQ=\angle RQO=90{}^\circ -75{}^\circ =15{}^\circ \]

\[\therefore \angle QOR=(180{}^\circ -2\times 15{}^\circ )\]

\[=180{}^\circ -30{}^\circ =150{}^\circ \]

\[\angle QSR=\frac{1}{2}\angle QOR\]

\[=75{}^\circ \]                     (Angle subtended on arc is half the angle subtended on centre)

\[\therefore \] In \[\Delta \text{ }SQR\]

\[\angle RQS=180{}^\circ -(\angle SRQ+\angle RSQ)\]

\[=180{}^\circ -(75{}^\circ +75{}^\circ )\]

\[\therefore \angle RQS=30{}^\circ \]