AM is perpendicular to the plane of the rhombus ABCD with a length of 8 cm. It is known

AM is perpendicular to the plane of the rhombus ABCD with a length of 8 cm. It is known that the distance from point M to line BC is 10 cm, angle B is 120 degrees. Perform additional constructions and find the distance from point M to line BC.

Since, by condition, the greater angle of the rhombus is 120, the angle ABK = 180 – 120 = 60.

The AKB triangle is rectangular, since the AK is perpendicular to the BC, then the angle BAK = 180 – 90 – 60 = 30.

The leg BK lies opposite an angle of 30, which means BK = AB / 2 = 8/2 = 4 cm.

Then AK ^ 2 = AB ^ 2 – BK ^ 2 = 64 – 16 = 48.

In a right-angled triangle AMK, according to the Pythagorean theorem, AM ^ 2 = MK ^ 2 – AK ^ 2 = 100 – 48 = 52.

The desired distance MO is the perpendicular from point M to the diagonal BD of the rhombus.

The ABD triangle is equilateral, since AD ​​= AB, and the angle between them is 60, then AO = AB * √3 / 2 = 8 * √3 / 2 = 4 * √3.

Then MO ^ 2 = AO ^ 2 + AM ^ 2 = (4 * √3) ^ 2 + 5 ^ 2 = 100.

MO = 10 cm.

Answer: From point M to segment BD 10 cm.