Question

Question 13
In the figure, O is the centre of the circle $\angle BCO={30}^{\circ }$. Find x and y.

Answer 1

Given, O is the centre of the circle and $\angle BCO={30}^{\circ }$.
In the given figure join OB and AC.

In $\Delta BOC$,
CO = BO [both are the radius of circle]
$\angle OBC=\angle OCB={30}^{\circ }$ [angles opposite to equal sides are equal]
$\angle BOC+\angle OBC+\angle OCB={180}^{\circ }$ [by angle sum property of a triangle]
$\angle BOC={180}^{\circ }-(\angle OBC+\angle OCB)$
$\angle BOC={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={120}^{\circ }$

We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\angle BOC=2\angle BAC$
$\angle BAC=\frac{{120}^{\circ }}{2}={60}^{\circ }$
Also, $\angle BAE=\angle CAE={30}^{\circ }$ [AE is an angle bisector of angle A]
$\Rightarrow \angle BAE=x={30}^{\circ }$

$In\Delta ABE,\angle BAE+\angle EBA+\angle AEB={180}^{\circ }$ [by angle sum property of a triangle]
$\Rightarrow {30}^{\circ }+\angle EBA+{90}^{\circ }={180}^{\circ }$
$\angle EBA={180}^{\circ }-({90}^{\circ }+{30}^{\circ })={180}^{\circ }-{120}^{\circ }={60}^{\circ }$
$Now,\angle EBA={60}^{\circ }$
$\Rightarrow \angle ABD+y={60}^{\circ }$
$\Rightarrow \frac{1}{2}\times \angle AOD+y={60}^{\circ }$ [in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.]
$\Rightarrow \frac{{90}^{\circ }}{2}+y={60}^{\circ }[\therefore \angle AOD={90}^{\circ }(given\left)\right]$
$\Rightarrow {45}^{\circ }+y={60}^{\circ }$
$\Rightarrow y={60}^{\circ }-{45}^{\circ }$
$\therefore y={15}^{\circ }$