In parallelogram ABCD, the diagonal AC is 2 times the side AB and the angle ACD = 74 degrees.

In parallelogram ABCD, the diagonal AC is 2 times the side AB and the angle ACD = 74 degrees. Find the angle between the diagonals of the parallelogram.

By the property of the parallelogram, its diagonal, at the intersection point, is divided in half, then, since AB = CD, and by condition, AC = 2 * AB, then AC = 2 * CD, and then CO = CD = AC / 2.

Since CO = CD, the triangle COD is isosceles, and then the angle COD = CDO = (180 – ACD) / 2 = (180 – 74) / 2 – 53.

Then the angle BOS = 180 – 53 = 127.

Answer: The angles between the diagonals of the parallelogram are 53 and 127.