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In the given figure, $O$ is the centre of the circle. If $\angle AOB = 145^\circ$, then find the value of $x$.
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Take a point $P$ on circumference and join $$\begin{aligned}& AP \& BP. \\ & \angle APB = \frac{1}{2} \times 145^\circ = 72.5^\circ \\ & \angle APB + \angle ACB = 180^\circ \\ & \Rightarrow \angle ACB = 107.5^\circ \text{ or } x = 107.5^\circ\end{aligned}$$