question_answer

P and Q are two points on a circle with centre at O. R is a point on the minor arc of the circle between the points P and Q. The tangents to the circle at the points P and Q meet each other at the point S. If \[\angle PSQ=20{}^\circ ,\] then \[\angle PRQ\] is equal to

A) \[200{}^\circ \]            

B) \[160{}^\circ \]

C) \[100{}^\circ \]

D) \[80{}^\circ \]

Correct Answer: C

Solution :

Join P and Q with an another point, say T, on the major arc.

Also, join PO and QO.

In POQS, \[\angle PSQ=20{}^\circ \]

\[\angle OPS=\angle OQS=90{}^\circ \]              [\[\because \]Tangent]

\[\angle POQ=360{}^\circ -(90{}^\circ +90{}^\circ +20{}^\circ )=160{}^\circ \]

\[\therefore \] \[\angle PTQ=\frac{1}{2}\angle POQ=\frac{1}{2}\times 160{}^\circ =80{}^\circ \]

Now, PTQR is a cyclic quadrilateral.

\[\angle PRQ=180{}^\circ -\angle PTQ\]

\[=180{}^\circ -80{}^\circ =100{}^\circ \]