Find the sides of a parallelogram if its diagonals are 40cm and 24cm, and the angle between the diagonals is 40 degrees.

The diagonals of the parallelogram are halved at the point of their intersection.

Then: ОВ = ОD = ВD / 2 = 24/2 = 12 cm, ОА = OC = АС / 2 = 40/2 = 20 cm.

In the triangle BОС, by the cosine theorem, we define the length of the side ВС.

In triangle AOB, by the cosine theorem, we define the length of the side AB.

AB ^ 2 = OB ^ 2 + AO ^ 2 – 2 * OB * AO * Cos40.

AB ^ 2 = 144 + 400 – 2 * 12 * 20 * 0.77.

AB ^ 2 = 544 – 369.6 = 174.4

AB = √174.4 ≈ 13.2 cm.

The AOB angle is adjacent to the BOC angle, then the BOC angle = 180 – 40 = 140.

Then, by the cosine theorem:

AB ^ 2 = OB ^ 2 + AO ^ 2 – 2 * OB * AO * Cos140.

AB ^ 2 = 144 + 400 – 2 * 12 * 20 * (-0.77).

AB ^ 2 = 544 + 369.6 = 913.6.

AB = √913.6 ≈ 30.2 cm.

Answer: The sides of the parallelogram are 30.2 cm and 13.2 cm.