Find the sides of a parallelogram if its diagonals are 22 m and 16 m, and one of the angles

Find the sides of a parallelogram if its diagonals are 22 m and 16 m, and one of the angles formed by them is 120 degrees.

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

Then: ОВ = ОD = ВD / 2 = 16/2 = 8 cm, ОА = OC = АС / 2 = 22/2 = 11 cm.

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

BC ^ 2 = OB ^ 2 + OC ^ 2 – 2 * OB * OC * Cos120.

BC ^ 2 = 64 + 121 – 2 * 8 * 11 (-1/2).

BC ^ 2 = 185 + 88 = 273.

BC = √273.

The angle AOB is adjacent to the angle BOC, then the angle AOB = 180 – 120 = 60.

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

AB ^ 2 = OB ^ 2 + AO ^ 2 – 2 * OB * AO * Cos60.

AB ^ 2 = 64 + 121 – 2 * 8 * 11 (1/2).

AB ^ 2 = 185 – 88 = 97.

AB = √97.

Answer: The sides of the parallelogram are √273 cm and √97 cm.