In the cube abcd-a1b1c1d1, E1 is known as the midpoint of a1d1, and the dihedral angle e1-ab-c is calculated Why is ∠ e1ad the plane angle of the dihedral angle e1-ab-c? Isn't its edge AB?

In the cube abcd-a1b1c1d1, E1 is known as the midpoint of a1d1, and the dihedral angle e1-ab-c is calculated Why is ∠ e1ad the plane angle of the dihedral angle e1-ab-c? Isn't its edge AB?

The plane angle of dihedral angle e1-ab-c is e1ad
tan(∠E1AD)=tan∠AE1A1=2
∠E1AD=arctan2

In cube ABCD with edge length 1_ In a1b1c1d1, find dihedral angle A_ B1D1_ The complementary value of C
Another question: find the dihedral angle C1 formed by the plane c1bd and the bottom ABCD_ BD_ The tangent value of the plane angle of C (hurry! Don't analyze

Find the intersection O1 of the bottom a1b1c1d1 diagonal and connect AO1, CO1 and AC; AO1 = CO1 = √ (1 ^ 2 + (√ 2 / 2) ^ 2) = √ 6 / 2; AC = √ 2; cos (∠ ao1c) = (2ao1 ^ 2-ac ^ 2) / (2. AO1. AO1) = 1 / 3; 2. Find the intersection o of the top ABCD diagonal and connect CO and C1O; CO = AC / 2 = √ 2 / 2; Tan (∠ COC

In the cube abcd-a1b1c1d1, the degree of dihedral angle a-bd1-a1 is______ ．

Taking a as the coordinate origin, AB, ad, Aa1 as the X, y, Z axis positive direction respectively, the space coordinate system is established. According to the structural characteristics of the cube, let the edge length of the cube be 1, we can easily get: Ab1 = (1, 0, 1) is a normal vector of the plane abd1, DA1 = (0, - 1, 1) is a normal vector of the plane bd1a1, let the degree of the dihedral angle a-bd1-a1 be θ, then cos θ = Ab1 · DA1 | Ab1 | · | DA1 | = 12 then θ =60 degree, so the answer is: 60 degree

In the cube abcd-a1b1c1d1, the tangent of the dihedral angle a1-bd-a formed by section a1bd and ground ABCD is equal to?
Root 3 / 3? Root 2 / 2? Root 2? Root 3?

=1