(1) Prove that there is a little E on the side AA 1, so that OE⊥ plane BB1c, and find the length of AE;
(2) Find the cosine of the angle between plane A 1B 1C and plane BB1c.
(1) Proof: ∵ In the triangular prism ABC-A 1 b1,it is known that AB=AC=AA 1=√5, BC=4, A1.
∴aa 1// surface bb 1c 1c = = >facea 1ao⊥face ABC = = & gt； Bc⊥a 1ao = = > aircraft; Face A 1AO⊥ face BB 1C 1C
O is OE⊥AA 1, AA 1 is E.
∴OE⊥ surface BB 1C 1C
Connect OA, OA = √ (ab 2-ob 2) = 1.
a 1o=√(aa 1^2-oa^2)=2
oa^2=ae*aa 1==>； AE=√5/5
(2) Analysis: Find the cosine of the included angle between plane A 1B 1C and plane BB1c.
The f area C 1F⊥B 1C spans C 1, and the g area fg ⊥ B 1C spans f, connecting GC 1.
∴∠GFC 1 is the plane angle between plane A 1B 1C and plane bb1c.
∵BB 1C 1C is a rectangle, ∴ cc 1b 1 = π/2.
At ⊿CB 1C 1, b1c = √ (b1c12+cc12) = √ 21.
b 1c 1^2=b 1f*b 1c==>； 4^2=b 1f*√2 1==>； B 1F= 16/√2 1
fc 1=√(b 1c 1^2-fb 1^2)=？ 4√5/√2 1?
From (1)A 1O=2, OC=2, ∴A 1C=2√2.
In ⊿A 1CB 1
cos∠a 1cb 1=(a 1c^2+b 1c^2-a 1b 1^2)/(2a 1c*b 1c)=(8+2 1-5)/(2*2√42)=6/√42
CF =√2 1- 16/√2 1 = 5/√2 1
tan∠a 1cb 1 = GF/CF =√6/6 = = & gt； GF=5√ 14/42？
cos∠a 1cb 1 = CF/CG = 6/√42 = = & gt； CG=5/√2 1*√42/6=5√2/6
In ⊿A 1CC 1
cos∠a 1cc 1=(a 1c^2+c 1c^2-a 1c 1^2)/(2a 1c*c 1c)=(8+5-5)/(2*2√ 10)=2/√ 10
CG=5√2/6
gc 1=√(gc^2+cc 1^2-2*gc*cc 1*cos∠a 1cc 1)=√(50/36+5-2*5√2/6*√5*2/√ 10)
=√(55/ 18)?
At ⊿GFC 1
cos∠gfc 1=(gf^2+fc 1^2-gc 1^2)/(2gf*c 1f)=(25/ 126+80/2 1-55/ 18)/(2*5√ 14/42*√80/√2 1)
=√30/ 10