∴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.