Solution (1)∵PC⊥ plane ABC, AB? 6? 3 aircraft ABC, ∴ PC ⊥ AB.

∵CD⊥ aircraft PAB, AB? 6? 3 aircraft Pabu, ∴ CD ⊥ AB.

PC∩CD=C, ∴AB⊥ flat PCB ..

(2) Take the midpoint e of AP and connect CE and DE.

∵PC=AC=2,∴CE⊥PA,CE= 2。

∵CD⊥ Aircraft Company,

De ⊥ pa is obtained from the inverse theorem of three vertical line theorems.

∴∠CED is the plane angle of dihedral angle c-pa-b.

From (1)AB⊥ flat PCB and ab = bc, BC= 2. Available.

In Rt△PCB, PB= PC2+BC2=6.

CD=PC？ 6? 1BCPB=2×26=23。

At Rt△CDE,

sin∠CED= CDCE=232=63。

∴ cos∠CED=33