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 \[\Delta \text{ }SQR\] \[\angle RQS=180{}^\circ -(\angle SRQ+\angle RSQ)\] \[=180{}^\circ -(75{}^\circ +75{}^\circ )\] \[\therefore \angle RQS=30{}^\circ \]
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