1, in triangular prism ABC-a1b1c1,aa 1∑bb 1.

Then ∠B 1BC or the rest corner is the demand.

In △ABC, AB⊥ and = AC, then BC=2*2?

Similarly, B 1C 1=2*2?

In the plane △A 1BC 1, a1b = a1= 2, and A 1B⊥ ABC.

So BC 1=2*2?

In △ABA 1, AB⊥ and = a1b.

So AA 1=2*2?

Prism ABC-a1b1c1.

aa 1 = bb 1 = cc 1 = 2 * 2？

In the quadrilateral BCC 1B 1

∫bb 1 = cc 1 = 2 * 2？ =BC=B 1C 1

The quadrilateral BCC 1B 1 is a diamond.

And diagonal BC 1=2*2? =BB 1=B 1C

∴∠ b1bc =120 > 90, that is, the angle between AA 1 and BC is 60.

2. Connect BC 1

∵A 1B⊥ surface ABC

∴A 1B⊥AB

∵AB⊥AC,AC∥A 1C 1

∴AB⊥A 1C 1

∵A 1c 1 cross a 1 b in a1.

∴AB⊥ Plane A 1BC 1

In △A 1B 1C 1, p is pm ∑ a1c and a1m.

In the surface ABB 1A 1, extend AB, span m is MN∑a 1B, span AB extends to n. connect PN.

∫PM∑a 1c 1, MN∑a 1b, pm intersects Mn in the plane of m, ab ⊥ a1BC.

∴AB⊥ PMN

∴∠MNP is the plane angle of dihedral angle P-AB-A 1

And A 1B⊥ plane A 1B 1C 1, Mn ∑ a1b.

Then MN⊥ Aircraft A 1B 1C 1

∴ triangle PMN is a right triangle, ∠ NMP = 90.

When cos∠MNP=(2/5)*5? And Mn Σ and = a1b = 2.

Then PM= 1.

∫2PM∑and = a 1c 1

P is the midpoint of B 1C 1.

That is, when p is at the midpoint of B 1C 1, the cosine of the dihedral angle P-AB-A 1 is (2/5)*5?