O is joined to points C and D, and B is joined to C. In triangle COD: OC = OD = CD = radius of the circle \(\therefore \angle\)COD = \(60^\circ\) Now, the arc CD subtends a central angle \(\angle\)COD and an inscribed angle \(\angle\)CBD \(\therefore \angle\)CBD = \(\frac{1}{2} \times \angle\)COD = \(\frac{1}{2} \times 60^\circ = 30^\circ\) \(\therefore \angle\)CBP = \(30^\circ\) Also, \(\angle\)ACB = \(90^\circ\) [angle in a semicircle] \(\therefore \angle\)BCP = \(180^\circ - \angle\)ACB = \(180^\circ - 90^\circ = 90^\circ\) Now, in triangle BCP: \(\angle\)CPB = \(180^\circ - (\angle\)BCP + \(\angle\)CBP) = \(180^\circ - (90^\circ + 30^\circ) = 180^\circ - 120^\circ = 60^\circ\) \(\therefore \angle\)APB = \(60^\circ\)