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 from the point S are drawn which touch the circle at P and Q. If \[\angle PSQ=20{}^\circ ,\]then \[\angle PRQ\]is equal to                                                                                      [SSC (CGL) 2013]

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 quadrilateral POQS,

\[\angle PSQ=20{}^\circ \]

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

\[\therefore \]                  \[\angle POQ=360{}^\circ \]

\[-\,\,(90{}^\circ +90{}^\circ +20{}^\circ )=160{}^\circ \]

\[\therefore \]      \[\angle PTQ=\frac{1}{2}\angle POQ\]

\[=\frac{1}{2}\times 160{}^\circ =80{}^\circ \]

Now, PTQR is a cyclic quadrilateral.

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

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