Answer: C
\(\angle\)POA = 120° ∴ In triangle BOP, \(\angle\)PBO + \(\angle\)BPO = exterior angle \(\angle\)POA i.e., 2\(\angle\)PBO = 120° [∵ OB = OP = radius of the circle ∴ \(\angle\)PBO = \(\angle\)BPO] Therefore, \(\angle\)PBO = 60°
\(\angle\)POA = 120° ∴ In triangle BOP, \(\angle\)PBO + \(\angle\)BPO = exterior angle \(\angle\)POA i.e., 2\(\angle\)PBO = 120° [∵ OB = OP = radius of the circle ∴ \(\angle\)PBO = \(\angle\)BPO] Therefore, \(\angle\)PBO = 60°