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

The correct option is B ${87}^{\circ }$

$\Delta QPR$ is an isosceles triangle with $QP=QR$

$\Rightarrow \angle QRS=\angle QPS=2^o$

$\Rightarrow \angle RQP=180-2-2={176}^o$

Also, in $\Delta PQS,PQ=PS=1$

$\Rightarrow \angle PQS=\angle PSQ=\frac{180-2}{2}={89}^o$

Now $\angle RQP=\angle RQS+\angle PQS$

$\therefore {176}^o=\angle RQS+{89}^o$

$\therefore \angle RQS={87}^o$