ABCD is a rhombus. angle A = 60 degrees, AB = m, BE perpendicular to ABC, BE = m√3 / 2.

ABCD is a rhombus. angle A = 60 degrees, AB = m, BE perpendicular to ABC, BE = m√3 / 2. Find the angle between the planes AED and ABC.

Since ABCD is a rhombus, then AB = AD = BC = CD = m cm.

The angle BAD, by condition, is equal to 60, then the triangle ABD is equilateral.

Let’s build the height BH of an equilateral triangle ABD. BH = AD * √3 / 2 = m * √3 / 2 cm.

The height of BH is the projection of the inclined EH onto the plane ABCD, and since BH is perpendicular to AD, then EH is perpendicular to AD.

Then the linear angle EHB is the desired angle between the planes.

The BEN triangle is rectangular, then tgBHE = BE / BH = (m * √3 / 2) / (m * √3 / 2) = 1.

Then the angle BHE = 45.

Answer: The angle between the planes is 45.