The cosine value of the dihedral angle b-dc '- C is? I take d 'as the origin to build the space right-handed system, and the result shows that the normal vector of DCC' is (0,0,0), which is unscientific!

The cosine value of the dihedral angle b-dc '- C is? I take d 'as the origin to build the space right-handed system, and the result shows that the normal vector of DCC' is (0,0,0), which is unscientific!

The upper be should be √ 6 / 2, not √ 2 / 2, it is positive △ c'bd side, the height of DC 'side should be √ 2 * √ 3 / 2 = √ 6 / 2. △ DCC' is the projection of △ BDC 'on plane dcc'd', s △ BDC '= √ 3 (√ 2) ^ 2 / 4 = √ 3 / 2, s △ DCC' = 1 / 2, let the dihedral angle b-dc '- C plane angle be θ, s △ BDC' * cos θ = s △ DCC '

In the cube abcd-a1b1cd1d1, find the cosine value of the dihedral angle b-a1c1-d
Map and process, the key is to map!

Connect a1c1 and b1d1 to point E
Because a1b1c1d1 is a square, a1c1 ⊥ b1e
BB1 vertical plane a1b1c1d1, b1e is the projection of be on the plane
So be ⊥ a1c1
Similarly, de ⊥ a1c1
So ∠ bed is the dihedral angle
Let the edge length of cube be a
Then b1e = √ 2A / 2, BB1 = a
So be = √ 6A / 2
Simple has △ bb1e ≌ △ dd1e, so de = be = √ 6A / 2
BD is a square ABCD diagonal, so BD = √ 2A
cos∠BED=（BE²+DE²-BD²）/（2BE×DE）
=(3a²-2a²)/(3a²)
=1/3