∴cd⊥pd.∵pd⊥ Aircraft ABCD

Cd⊥pd ∵cd⊥ad, PD ∩ ad = d, ∴pa⊥cd.∴cd⊥ flat pad.

From PA⊥DE, PA⊥CD, de ∩ CD = d, we get: PA⊥ plane EFCD.

(2)

∵PD⊥ Plane ABCD, ∴PD⊥AD, and AD = √ 2, PD = 2, ∴ PA = √ (AD=√2+PD=2) = √ 10.

∵DE⊥PA、pd⊥ad,∴de pa = PDad,∴√ 10de=2√2,∴de=2/√5。

According to the projective theorem, PE PA = PD 2, ∴√10pe = 4, ∴ PE = 4/√ 10.

∵CD⊥ flat pad, ∴DE⊥CD, ∴ ce 2 = de 2+CD 2,

∵PA⊥ plane EFCD, ∴PE⊥CE, ∴ PC 2 = PE 2+Ce 2 = PE 2+de 2+CD 2,

DE = 2/√ 5，CD = 4，∴PC 2 = 16/ 10+4/5+ 16 = 12/5+ 16 = 92/5。

∵PE⊥ plane EFCD, ∴∠ PCE = the angle formed by PC and plane EFCD.

And sin ∠ PCE = PE/PC = (4/√10)/(4 √ 23/√ 5) = √ 46/46.

The sine value of the angle between PC and EFCD plane is ∴ 46/46.