∵∠APB＝60°,PA、PB切⊙O于A、B
∴△APB为正三角形,∠AOB=120°
∴OE=1/2OA,AB=AP=4
又∵PA²=PC*PD,PD＝3PC,PA=4
∴PC=4√3/3,PD=4√3
∴OA=(PD-PC)/2=4√3/3
S-ACB=S扇形OACB-S△ABO
=π*OA²*120°/360°-1/2AB*OE
=（16π-12√3）/9
图在这里http://hi.baidu.com/%D2%D7%CB%AE%D0%A1%D9%E2/album/item/5277d94443a9822693b3636c8a82b9014b90eb44.html