In the rhombus ABCD, the diagonals intersect at point O, A = 31. Find the angles of the triangle BOC.

By the property of a rhombus, the sum of its adjacent angles is 180.

Angle ABC + ВAD = 180, then angle ABC = 180 – ВAD = 180 – 31 = 149.

The opposite angles at the rhombus are equal, then the ВСD angle = ВAD = 31.

By the property of the diagonals of the rhombus, they divide the angles at the vertices of the rhombus in half, and intersect at right angles.

Then, in the ВOС triangle, the angle ВOС = 90, the angle ВCO = ВСD / 2 = 31/2 = 15.5, the angle СВO = ABC / 2 = 149/2 = 74.5.

Answer: The angles of the AOB triangle are 15.5, 74.5, 90.