A chord of a circle is equal to its radius. The angle subtended by this chord at a point on the circumference in the major segment is

∴ OA = OB = AB

∴ ΔOAB is an equilateral triangle.

∴ ∠AOB = 60

The angle subtended by an arc at the circumference is half of that at the centre.

$\therefore \angle \mathrm{ACB}=\frac{1}{2} \times 60^{\circ}=30^{\circ}$