Question

In the figure given 'O' is the centre of the circle. If QR$=$OP and $\angle$ORP$={20}^o$. Find the value of 'x' giving reasons.

Answer 1

$QR=OP⟹QR=OQ$ (as $OP=OQ$ are radii of circle)

$\therefore \angle ROQ=\angle ORP={20}^{\circ }$

Hence, using sum of interior angles theorem, $\angle ORQ={180}^{\circ }-\angle ROQ-\angle ORP={140}^{\circ }$

$\therefore \angle OQP={180}^{\circ }-\angle OQR={40}^{\circ }$

Also, $OP=OQ$ as both are radii of the same circle.

$\therefore \angle OQP=\angle OPQ={40}^{\circ }$ (isoceles triangle property)

$\therefore \angle POQ={180}^{\circ }-\angle OQP-\angle OPQ={100}^{\circ }$ (sum of interior angles of a triangle)

Now,

$\angle TOP+\angle POQ+\angle QOR={180}^{\circ }$ (angles on a straight line)

$\therefore \angle TOP={180}^{\circ }-{100}^{\circ }-{20}^{\circ }={60}^{\circ }$

This is the required solution.