Mathematics

In the given figure, $$O$$ is the centre of a circle in which $$\angle OAB=20^o$$ and $$\angle OCB=55^o$$. Find 
$$\angle AOC$$


SOLUTION
We know that $$OA=OB$$ which is the radius 

The base angles of an isosceles triangle are equal

So we get

$$\angle OBA =\angle OAB =20^o$$

In $$\triangle AOB$$

Using the angle sum property

$$\angle AOB + \angle OAB +\angle OBA =180^o$$

By substituting values

$$\angle AOB +20^o +20^o =180^o$$

On further calculation 

$$\angle AOB =180^o -20^o -20^o$$

By subtraction 

$$\angle AOB =180^o -40^o$$

So we get

$$\angle AOB =140^o$$

We know that

$$\angle AOC =\angle AOB -\angle BOC$$

By substituting the values 

$$\angle AOC =140^o -70^o$$

So we get

$$\angle AOC =70^o$$
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