Solution: let ∠ POC = a (0

Connecting AB, it is easy to know that b+a/2=60 degrees, ∴ B = 60-A/2, ∴ 30.

In triangular POC, according to sine theorem | PC | = Sina | op |/sinb = Sina/sinb,

Therefore, the vector op * PC = | op || PC | cos < OP, PC> = Sina/sinb * cos (a+b) = sin (120-2b) * cos (120-2b+b)/sinb.

=sin(3b-60)/(2sinb)- 1/2 (according to the sum-difference product formula)

This problem is transformed into the problem of finding the range ∫30 of this function.