OA = OB, since both are radii. Thus triangle AOB is isosceles and \(\angle OAB=28^\circ\). The angles of a triangle sum to \(180^\circ\), so \(\angle AOB=180^\circ-28^\circ-28^\circ=124^\circ\). The angle at the circumference is half the angle at the centre standing on the same arc AB, hence \(\angle ACB=62^\circ\).