question_answer

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,              \[\angle PRQ=120{}^\circ \]