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In the cube abcd-a'b'c'd ', find the size of the dihedral angle B' - BD '- C'
Connect a 'C', intersect B'd 'with E, then c' e ⊥ B'd ', and BB' ⊥ plane a 'B' C'd ', C' e ∈ plane a 'B' C'd ', BB' ⊥ C 'e, BB' ∩ B'd '= B' ∩ C 'e ⊥ plane BB'd', connect be, then △ BB 'e is the projection of △ BC' E on plane BB'd ', let the dihedral angle B' - BD '- C'
In the cube ABCD-A "B" C "D, the size of dihedral angle B-A" C-A "is?
Let the edge length of the cube be a
It is known that △ a "BC is RT △ and a" C = √ 3 A, a "B = √ 2 A, BC = a, ∠ a" BC = 90 °
Take the midpoint o of AC,
Even Bo, then Bo ⊥ AC
In the plane ABCD ⊥ plane a "Ca, the intersection of these two sides is AC, and point B is in the plane ABCD,
The projection of point B in plane a "CA is point o
The projection of △ a "BC in plane a" CA is △ a "OC
If we set the dihedral angle B-A "and the size X of C-A,
Then (△ a "BC area) × cosx = (△ a" OC area)
Area S1 of ∵ △ a "BC = (1 / 2) × a" B × BC
= （1/2）× （√2 a） × a
=Square of (√ 2 / 2) a
The area S2 of ∵ △ a "OC = (1 / 2) × a" a × OC
= （1/2） × a ×（√2/2 a）
=Square of (√ 2 / 4) a
∴ cosx = S2 / S1
=[(√ 2 / 4) a squared] / [(√ 2 / 2) a squared]
= 1 / 2
X = 60 ° or π / 3
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