In the given figure, O is the centre of the circle. If $\displaystyle \angle \mathrm{AOB}=145^{\circ}$, then find the value of x.

Take a point P on circumference and join AP & BP. \[\begin{aligned} & \angle \mathrm{APB}=\frac{1}{2} \times 145^{\circ}=72.5^{\circ} \\ & \angle \mathrm{APB}+\angle \mathrm{ACB}=180^{\circ} \\ & \Rightarrow \angle \mathrm{ACB}=107.5^{\circ} \text { or } \mathrm{x}=107.5^{\circ} \end{aligned} \]