Question

In the given figure, $O$ is the centre of circle. If $\angle AOD={140}^o$ and $\angle CAB={50}^o$, calculate
$\angle EDB$

Answer 1

We know that

$\angle BOD+\angle AOD={180}^o$

By substituting the values

$\angle BOD+{140}^o={180}^o$

On further calculation

$\angle BOD={180}^o-{140}^o$

By subtraction

$\angle BOD={40}^o$

We know that $OB=OD$

So we get $\angle OBD=\angle ODB$

Consider $△OBD$

By using the angle sum property

$\angle BOD+\angle OBD+\angle ODB={180}^o$

We know that $\angle OBD=\angle ODB$

So we get

${40}^o+2\angle OBD={180}^o$

On further calculation

$2\angle OBD={180}^o-{40}^o$

By subtraction

$2\angle OBD={140}^o$

By division

$\angle OBD={70}^o$

We know that $ABCD$ is a cyclic quadrilateral

$\angle CAB+\angle BDC={180}^o$

$\angle CAB+\angle ODB+\angle ODC={180}^o$

By substituting the values

${50}^o+{70}^o+\angle ODC={180}^o$

On further calculation

$\angle ODC={180}^o-{50}^o-{70}^o$

By subtraction

$\angle ODC={180}^o-{120}^o$

So we get

$\angle ODC={60}^o$

Using the angle sum property

$\angle EDB+\angle ODC+\angle ODB={180}^o$

By substituting the values

$\angle EDB+{60}^o+{70}^o={180}^o$

On further calculation

$\angle EDB={180}^o-{60}^o-{70}^o$

By subtraction

$\angle EDB={180}^o-{130}^o$

So we get

$\angle EDB={50}^o$