Mathematics

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


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

The base angles of an isosceles triangle are equal

So we get

$$\angle OBC =\angle OCB =55^o$$

In $$\triangle BOC$$

Using the angle sum property

$$\angle BOC + \angle OCB +\angle OBC =180^o$$

By substituting values

$$\angle BOC +55^o +55^o =180^o$$

On further calculation 

$$\angle BOC =180^o -110^o$$

So we get

$$\angle BOC =70^o$$
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