In rhombus ABCD, the diagonal AC is twice as short as the diagonal BD. Find the tangent of the angle BAC.

Let the length of the diagonal BD = X cm, then, by condition, the length of the diagonal AC = BD / 2 = X / 2.

The diagonals of the rhombus, at the point of their intersection, are divided in half, then the length of the segment BO = BD / 2 = X / 2, and the length of the segment AO = AC / 2 = (X / 2) / 2 = X / 4.

Since the diagonals of the rhombus intersect at right angles, the triangle AOB is rectangular.

In a right-angled triangle, the tangent of the angle is equal to the ratio of the opposite leg to the adjacent leg.

tgbao = tgbac = BO / AO = (X / 2) / (X / 4) = 2.

Answer: The tangent of the angle BAC is 2.