Mathematics

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$$


SOLUTION
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 sun property

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

By substituting the values

$$ABC +100^o +50^o +=180^o$$

On further calculation

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

By subtraction

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

So we get

$$\angle ABC =30^o$$ 

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