A math problem in Hong Kong senior high school, geometry, finding the included angle. I want to ask how to do 2 1(b).

B, ① connect AE,∵DC⊥AC,DC⊥CE,∴DC⊥ plane ACE, and plan ABDC⊥ plane ACE;

∵ab∨CD, ∴AB⊥ Plane ACE, then AB⊥AE,⊿ABE are right triangles.

∵AB=6, BE= 10, ∴AE=8, CE=DF=8, then ⊿ACE is an isosceles triangle.

Let the high eh at the bottom of ⊿ACE get AH=HC=AC/2=BD/2=2, EH=√(8? -2? )=2√ 15。

ABDC aircraft ACE, EH⊥AC aircraft, eh ⊥ aircraft abdc.

BH is the projection of BE on the ABDC plane, and ∠EBH is the intersection angle between BE and ABDC plane.

In the right triangle EHB, ∫ sinebh = eh/be = 2 √15/10 = √15, ∴∠EBH=50.77.

The included angle between beryllium and ABDC plane is about 50.77 degrees.