# The diagonal of the rectangle forms an angle of 44 degrees on one of the sides.

**The diagonal of the rectangle forms an angle of 44 degrees on one of the sides. Find the angle between the diagonals of the rectangle and the angle that forms the diagonal on the other side.**

Let’s designate this rectangle ABCD. AC and BD are its diagonals.

The diagonals of the rectangle intersect at point O and the intersection point is halved. Thus, four isosceles triangles are obtained:

∆АОВ = ∆СОD;

∆AOD = ∆BOC.

Consider a triangle ∆AOD. Since the angles at the base of this triangle are 44 °, and the sum of all the angles of the triangle is 180 °, then:

∠AOD = 180 ° – ∠OAD – ∠ODA;

∠AOD = 180 ° – 44 ° – 44 ° = 92 °.

Consider a triangle ∆AVD. ∠WATER is a flat angle that is 180 °. In this way:

∠BOA = 180 ° – AOD;

∠BOA = 180 ° – 92 ° = 88 °.

Consider the triangle ∆AOB. Since the angle ∠BOA is 88 °, and the angles at the base are equal, then:

∠BAO = (180 ° – 88 °) / 2 = 92 ° / 2 = 46 °.

Answer: the angles between the diagonals are 92 ° and 88 °, the angle between the diagonal and the other side is 46 °.