Question

In the given figure, $AB$ and $CD$ are straight lines through the
centre $O$ of a circle. If $\angle AOC={80}^o$ and $\angle CDE={40}^o$, find $\angle ABC$

Answer 1

We know that $\angle AOC$ and $\angle BOC$ from a linear pair

It can be written as

$\angle BOC={180}^o-{80}^o$

By subtraction

$\angle BOC={100}^o$

Using the angle sum property

$\angle ABC+\angle BOC+\angle DCE={180}^o$

By substituting the values

$\angle ABC+{100}^o+{50}^o+={180}^o$

On further calculation

$\angle ABC={180}^o-{100}^o-{50}^o$

By subtraction

$\angle ABC={180}^o-{150}^o$

So we get

$\angle ABC={30}^o$