Question

In the figure above $P,Q,R$ and $S$ are concyclic points, $O$ is the centre of the circle $PQ=QR=RS$. If $\angle OPS={15}^{\circ }$ then find $\angle OQR$

• A. ${50}^{\circ }$
• B. ${55}^{\circ }$
• C. ${60}^{\circ }$
• D. ${65}^{\circ }$

Answer 1

The correct option is D ${65}^{\circ }$

Solution:

Given that:

$\angle OPS={15}^{\circ }$ and PQ=QR=RS

To find:$\angle OQR=?$

Solution:

Since O is the centre, $OP=OS=r$ $\therefore \angle OPS=\angle OSP={15}^{\circ }$

where, r is the radius of the circle.

In $△OPS$

$\angle OPS+\angle OSP+\angle POS={180}^{\circ }$

or, ${15}^{\circ }+{15}^{\circ }+\angle POS={180}^{\circ }$

or, $\angle POS={150}^{\circ }$

Again, since $PQ=QR=RS\therefore \angle POQ=\angle QOR=\angle ROS$

$\angle POQ+\angle QOR+\angle ROS=\angle POS={150}^{\circ }$

or, $3\angle QOR={150}^{\circ }$

or, $\angle QOR={50}^{\circ }$

Again, since $OQ=OR=r\therefore \angle OQR=ORQ$

In $△OQR$

$\angle OQR+\angle ORQ+\angle QOR={180}^{\circ }$

or, $2\angle OQR+{50}^{\circ }={180}^{\circ }$

or, $\angle OQR={65}^{\circ }$

Hence, D is the correct answer.