Question

In the given figure, O is the centre of the circle .What is the value of $x?$

• A. ${115}^o$
• B. ${125}^o$
• C. ${155}^o$
• D. ${230}^o$

Answer 1

The correct option is B ${125}^o$

Given: $O$ is the centre of the circle.

So, $OP,OR,OQ$ are the radius of the circle.

So, $OP=OR=OQ$

Now,

Since $OP=OQ$

$\therefore △OPQ$ is isosceles.

Hence $\angle OPQ=\angle OQP={25}^{\circ }$

From angle sum property,

$\angle POQ={180}^{\circ }-(\angle OPQ+\angle OQP)$

$={180}^{\circ }-({25}^{\circ }+{25}^{\circ })={180}^{\circ }-{50}^{\circ }={130}^{\circ }$

Again, $OQ=OR$

$\therefore △OQR$ is isosceles.

Hence $\angle ORQ=\angle OQR={30}^{\circ }$

By angle sum property

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

$={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

We know that, in a circle the central angle is twice of any inscribed angle subtended by the same arc.

Here, $\angle POR$ is the central angle and $\angle PSR$ is the inscribed angle subtended by the same arc.

$\therefore \angle POR=2\angle PSR$

$\Rightarrow \angle POQ+\angle QOR=2\angle PSR$

$\Rightarrow {130}^{\circ }+{120}^{\circ }=2x$

$\Rightarrow 2x={250}^{\circ }$

$\Rightarrow x={125}^{\circ }$