The plane proof of three-dimensional geometric space in senior two mathematics.

Let the vertical line of BD pass through A 1, A 1E = 3× 4/5 = 2.4,

The other angle A 1-BD-C is a straight dihedral angle, so A 1E is perpendicular to the curved surface BCD.

A 1E is perpendicular to BC.

E is taken as the vertical line EF of BC, because BC is perpendicular to A 1E and EF respectively.

So BC is perpendicular to the surface A 1EF. Connect A 1F, then BC is perpendicular to a1f.

Angle A 1FE is a dihedral angle.

Because DE 2 = AD 2-A 1e 2, de =1.8; BE = 3.2EF:CD=BE:BD

So EF=BE×CD/BD=3.2*4/5=2.56.

The tangent of the angle A 1-BC-D is a1e/ef =15/16.