Question

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

Answer 1

From the figure we know that $\angle AOD$ and $\angle OEC$ from right angles

So we get

$\angle AOD=\angle OEC={90}^o$

We know that $OD||BC$ and $OC$ is a transversal

From the figure we know that $\angle AOD$ and $\angle OEC$ are corresponding angles

$\angle AOD=\angle OEC$

We know that $\angle DOC$ and $\angle OCE$ are alternate angles

$\angle DOC=\angle OCE={30}^o$

Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

$\angle DOC=2\angle DBC$

It can be written as

$\angle DBC=\frac{1}{2}\angle DOC$

By substituting the values

$\angle DBC=\frac{30}{2}$

By division

$y=\angle DBC={15}^o$

In the same way

$\angle ABD=\frac{1}{2}\angle AOD$

By substituting the values

$\angle ABD=\frac{90}{2}$

By division

$\angle ABD={45}^o$

We know that

$\angle ABE=\angle ABC=\angle ABD+\angle DBC$

So we get

$\angle ABE=\angle ABC={45}^o+{15}^o$

By addition

$ABE=\angle ABC={60}^o$

Consider $△ABE$

Using the angle sum property

$\angle BAE+\angle AEB+\angle ABE={180}^o$

By substituting the values

$x+{90}^o+{60}^o={180}^o$

On further calculation

$x={180}^o-{90}^o-{60}^o$

By subtraction

$x={180}^o-{150}^o$

So we get

$x={30}^o$

Therefore , the value of $x$ is ${30}^o$ and $y$ is ${15}^o$.