In parallelogram ABCD, the diagonal AC is 2 times the side AB and ∠ACD = 104∘.

In parallelogram ABCD, the diagonal AC is 2 times the side AB and ∠ACD = 104∘. Find the acute angle between the diagonals of the parallelogram.

Let’s designate the point of intersection of the parallelogram diagonals through O. And since the diagonals intersect each other in the middle and AC = 2 * AB, then AB = AO = OC. The opposite sides of the parallelogram are also equal to each other (AB = CD).

We get a triangle OCD, in which OC = CD (isosceles triangle). The angles at the base of an isosceles triangle are equal. Let’s find them:

(180 – 104) / 2 = 38;

Answer: the angle COD (the angle between the diagonals of the parallelogram) is 380.