∴AB=2 times the root number 2

AP = BP = AB again.

∴BP=2 times the root number 2

PC = BC = 2

It can be concluded that PC2+BC2=BP2=8.

From Pythagorean Theorem, it can be concluded that △PCB is an isosceles right triangle.

That is ∠ PCB = 90, that is, PC⊥BC.

PC⊥AC, AC and BC are straight lines intersecting on the plane ABC.

∴PC⊥ Plane ABC

∴ any straight line in the vertical plane ABC of PC; AB∈ plane ABC

∴PC⊥AB

(2)△PCA passes through AP to D through the vertical line of point C, as shown in figure?

From known conditions, it can be concluded that point D is the midpoint of AP, and from known conditions, it can be concluded that △ABP is an equilateral triangle. If BD is connected, it can be concluded that BD is a vertical line of △ABP, which is BD⊥AP.

∴∠BDC is the plane angle of dihedral angle B-AP-C.

∵AC=PC=2, △ACP is an isosceles right triangle.

∴CD=AD= root number 2

Ab = ap = bp = twice the root number 2, BD⊥AP.

∴BD= is the square root of 6, BC =2.

All three sides of a triangle are known, and the dihedral angle can be obtained by sine or cosine.