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. |
\[\angle PRQ=180{}^\circ -\angle PTQ\] |
\[=180{}^\circ -80{}^\circ =100{}^\circ \] |
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