Question

In the figure, PQ = RS and $\angle ORS={48}^o$. Find $\angle OPQ$ and $\angle ROS$.

Answer 1

In given figure , PQ=RS and $\angle ORS={48}^{\circ }$.

In given figure O is the center of circle and point P,Q,R and S on the circumference of circle

Then line OP, OQ ,OR and OS are the radius of circle

So OP=OQ=OR=OS

In $\Delta OPQ$ and $\Delta ORS$

$PQ=RS$ (Given )

$OP=OR$ (Radius of circle)

$OQ=OS$ (Radius of circle)

$\therefore \Delta POQ\cong \Delta ORS$ by SSS congruency

$\angle ORS=\angle OPQ={48}^{\circ }$

$OR=OS$ (radius)

$\therefore \angle OSR=\angle ORS={48}^{\circ }$

In $\Delta ORS$

$\angle ROS+\angle ORS+\angle OSR={180}^{\circ }$

$\Rightarrow \angle ROS+{48}^{\circ }+{48}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ROS={180}^{\circ }-{48}^{\circ }-{48}^{\circ }={84}^{\circ }$

$\angle OPQ={48}^{\circ },\angle ROS={84}^{\circ }$