Question

In the given figure, $O$ is the centre of the circle. If $\angle AOB={140}^0$ and $\angle OAC={50}^0$; find:
$\angle ACB$,

Answer 1

Here $\angle ACB=\frac{1}{2}$Reflec of $\angle AOB$

$=\frac{1}{2}({360}^{\circ }-{140}^{\circ })={110}^{\circ }$