The diagonal of a rectangular box ABCDA1B1C1D1, the base of which is a square

The diagonal of a rectangular box ABCDA1B1C1D1, the base of which is a square, is twice the side of the base. Find the angles between the diagonals of the box.

Let us denote the side of the square at the base through X cm, AB = BC = CD = AD = X cm.

Then, by condition, the diagonal DВ1 of the parallelepiped is equal to 2 * X cm.

Consider a right-angled triangle ABD, whose leg AB = AD = X, then, by the Pythagorean theorem, BD ^ 2 = AB ^ 2 + AD ^ 2 = 2 * X ^ 2.

BD = √2 * X.

Consider a right-angled triangle BB1D.

Let’s define the angle ВDВ1 between the diagonals ВD and DВ1.

CosBDB1 = BD / DB1 = √2 * X / 2 * X = √2 / 2.

Angle ВDВ1 = arcos (√2 / 2) = 45.

Answer: The angle between the diagonals is 45.