From the fact that the projection E of condition point A on the diameter BC is the midpoint of OC, it is easy to get OE =12oa;

∴∠aop=60；

By OA=OB again? ∠ABP=30。

In Rt△AOP, because OA=2 and ∠ AOP = 60, the available AP = 23,

Can PB be obtained from the tangent line theorem? PC=AP2= 12。

So the answer is: 30, 12.