In figure 2, PQ is a chord of a, circle with centre O and PT is a tangent If \[\angle QPT=60{}^\circ \], find \[\angle PRQ\]. |
Answer:
Given, O is the centre of the given circle \[\therefore \] OQ and OP are the radius of circle. \[\because \] PT is a tangent \[\therefore OP\bot PT\] So, \[\angle OPT=90{}^\circ \] \[\therefore \angle OPQ=90{}^\circ -\angle QPT\] \[\angle OPQ=90{}^\circ -60{}^\circ \] [Given, \[\angle QPT=60{}^\circ \]] \[\angle OPQ=30{}^\circ \] \[\therefore \angle OQP=30{}^\circ \] [\[\because \,\,\Delta \text{ }OPQ\] is isosceles triangle] Now, in \[\Delta \,OPQ\] \[\angle POQ+\angle OPQ+\angle OQP=180{}^\circ \] \[\angle POQ+30{}^\circ +30{}^\circ =180{}^\circ \] \[\angle POQ=120{}^\circ \] reflex\[\angle POQ=360{}^\circ -120{}^\circ =240{}^\circ \] \[\therefore \angle PRQ=\frac{1}{2}\] reflex \[\angle POQ\] [\[\because \] The angle substended by an arc of a circle at the centre is double the angle substended by it at any point on the remaining part of the circle] \[\angle PRQ=\frac{1}{2}\times 240{}^\circ \] Hence, \[\angle PRQ=120{}^\circ \]
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