In figure, $O$ is the centre of the circle. If $\angle OAB=40^o$ and $\angle OCB=30^o$, find $\angle AOC$.

We know that sides are equal to the radius

So we get

From the figure we know that the base angles of an isosceles triangle are equal $\angle OBC=\angle OCB$

It is given that

So we get $\angle OBC=30^o....(1)$

Consider $△BOA$

We know that the sides are equal to the radius 

So we get

From the figure, we know that the base angles of an isosceles triangle are equal

It is given that 

So we get

So we get $\angle OBA=40^o....(2)$

We know that

By substituting the values 

So we get

We know that angle subtended by an arc of a circle at the center is double the angle subtended by the arc at any point on the circumference.

It can be written as

So we get

By multiplication 

Therefore, $\angle AOC=140^o$