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 $\angle QRP=2^o$. Let S be the point where the segment RP intersects the given circle. Then measure of $\angle RQS$ equals

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

Answer 1

The correct option is B ${87}^o$

$\angle RQP={176}^o$
$\angle SPQ=2^o$
$\angle SQP={89}^o(SP=PQ)$
$\angle SQR=176-89={87}^o$