Question

In given figure, O is the centre of the circle $\angle BCO={30}^{\circ }$ Find x and y

Answer 1

$O$ is the center of the circle and $\angle BCO={30}^o.$

In given figure join $OB$ and $AC.$

In $△BOC,$

$\Rightarrow$ $CO=BO$ [ Radius of a circle ]

$\therefore$ $\angle OBC=\angle OCB={30}^o$ [ Angles opposite to equal sides are equal ]

$\Rightarrow$ $\angle BOC+\angle OBC+\angle OCB={180}^o$ [ By angles sum property of a triangle ]

$\Rightarrow$ $\angle BOC+{30}^o+{30}^o={180}^o$

$\Rightarrow$ $\angle BOC={180}^o-{60}^o$

$\Rightarrow$ $\angle BOC={120}^o$

We know that, in a circle, the angle subtended by an arc at the center is twice the angle subtended by it at the remaining part of the circle.

$\Rightarrow$ $\angle BOC=2\angle BAC$

$\Rightarrow \angle BAC=\frac{{120}^o}{2}={60}^o$

$\Rightarrow$ $\angle BAE=\angle CAE={30}^o$ [ $AE$ is an angle bisector of $\angle A$ ]

$\Rightarrow$ $\angle BAE=x={30}^o$

In $△ABE,$

$\Rightarrow$ ${30}^o+\angle EBA+{90}^{=}{180}^o$

$\Rightarrow$ ${120}^o+\angle EBA={180}^o$

$\Rightarrow$ $\angle EBA={60}^o$

$\Rightarrow$ $\angle ABD+y={60}^o$

In a circle, the angle subtended by an arc at the center is twice the angle subtended by it at the remaining part of the circle.

$\Rightarrow$ $\frac{1}{2}\times \angle AOD+y={60}^o$

$\therefore$ $\frac{{90}^o}{2}+y={60}^o$

$\Rightarrow$ ${45}^o+y={60}^o$

$\Rightarrow$ $y={60}^o-{45}^o$

$\therefore$ $y={15}^o$