# ABCD parallelogram AB = 20 cm, angle BAD = 45 degrees, BM perpendicular to area ABCD, angle between MA

**ABCD parallelogram AB = 20 cm, angle BAD = 45 degrees, BM perpendicular to area ABCD, angle between MA and ABC 60 degrees. Find the distance from point M to ABC; dihedral angle MADB.**

Since MB is perpendicular to the plane ABCD, the triangle ABM is rectangular, then tgABM = BM / AB.

BM = AB * tgABM = 20 * tg60 = 20 * √3 cm.

Let’s define the dihedral angle MADB. The segment AB is the projection of the oblique AM, the segment BD is the projection of the oblique DM, then the value of the dihedral angle is equal to the value of the linear angle ABD.

The ABD angle can be determined if the ABCD rhombus, then the ABC angle = 180 – 45 = 135.

The diagonals of the rhombus are the bisectors of the angles at its vertices, then the angle ABD = 135/2 = 67.5.

Answer: From point M to the plane 20 * √3 cm, the dihedral angle is 67.5.