Solution: Let A be the symmetry point D of OB, connect CD to OB to P, connect AP, and let D be DN⊥OA to N, then the value of PA+PC is the smallest at this time.

∫ rt △ the coordinate of OAB vertex A is (9,0),

∴OA=9，

∫tan∠BOA = 33，

∴AB=33,∠B=60，

∴∠AOB=30，

∴OB=2AB=63，

According to the triangle area formula, s △ OAB =12× OA× ab =12× ob× am, that is, 9×33=63AM.

∴AM=92，

∴AD=2×92=9，

∠∠AMB = 90，∠B=60，

∴∠BAM=30，

∫∠BAO = 90，

∴∠OAM=60，

∵DN⊥OA，

∴∠NDA=30，

∴AN= 12AD=92, which is obtained by Pythagorean theorem: DN=AD2? AN2