∵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.