Answer: A
\(\angle\)AOP = 108°
∴ \(\angle\)BOP = 180° - 108° = 72°
∴ In \(∆BOP\), \(\angle\)BPO + \(\angle\)OBP = Exterior angle \(\angle\)AOP
or, \(2\angle\)BPO = 108°
[∵ OB = OP = Radius of the circle ∴ \(\angle\)BPO = \(\angle\)OBP]
or, \(\angle\)BPO = 54°.
\(\angle\)AOP = 108°
∴ \(\angle\)BOP = 180° - 108° = 72°
∴ In \(∆BOP\), \(\angle\)BPO + \(\angle\)OBP = Exterior angle \(\angle\)AOP
or, \(2\angle\)BPO = 108°
[∵ OB = OP = Radius of the circle ∴ \(\angle\)BPO = \(\angle\)OBP]
or, \(\angle\)BPO = 54°.