$\mathrm{AB}$ is a chord of a circle with centre $\mathrm{O}$. $\mathrm{C}$ is a point on the circumference of the circle in the minor sector. If $\angle A B O=$ $40^{\circ}$, what is the measure (in degree) of $\angle A C B$ ?

130°

Here, \(\angle\)ABO = \(\angle\)BAO (angles formed by the radius are equal)

In \(\Delta \)AOB, \(\angle\)ABO + \(\angle\)BAO + \(\angle\)BOA = \({180}^\circ\)

⇒ \(\angle\)BOA = \({180}^\circ\) - (40 + 40) = \({100}^\circ\)

⇒ \(\angle\)BOA (external) = \({360}^\circ\) - \({100}^\circ\) = \({260}^\circ\)

\(\angle\)ACB = \(\frac{1}{2}\) \(\angle\)BOA(external) (since angle formed by the arc at the center is double compared to angle formed by the arc at other points on the circumference).

Therefore, \(\angle\)ACB = \(\frac{1}{2}\) x \({260}^\circ\) = \({130}^\circ\)

Therefore, \(\angle\)ACB = \({130}^\circ\)