Question

Suppose $Q$ is a point on the circle with centre $P$ and radius $1$ , as shown in the figure, $R$ is a point outside the circle such that $QR=1$ and $\angle QRP=2^{\circ }$. Let $S$ be the point where the segment $RP$ intersects the given circle. Then, measure of $\angle RQS$ equals

• A. $86^{\circ }$
• B. $87^{\circ }$
• C. $88^{\circ }$
• D. $89^{\circ }$

Answer 1

Answer: (B)

$\angle QRP=2^{\circ }$
$PQ=QR=1$
$\therefore \angle QPR=2^{\circ }$
$\angle RQP=180^{\circ }-4^{\circ }=176^{\circ }$
$SP=SQ$ radii of circle
$\therefore \angle SQP=\angle QSP$
$=\frac{180^{\circ }-2^{\circ }}{2}$
$=\frac{178^{\circ }}{2}=89^{\circ }$
$\angle RQS=\angle RQP-\angle SQP$
$=176^{\circ }-89^{\circ }=87^{\circ }$