In the ABCD rhombus, the AC and BD diagonals intersect at the point O, the angle D = 120 degrees.

In the ABCD rhombus, the AC and BD diagonals intersect at the point O, the angle D = 120 degrees. Find the angles of the BOC triangle.

By the property of a rhombus, the sum of its adjacent angles is 180, then BCD + ABC = 180.

BCD = 180 – ABC = 180 – 120 = 60.

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 BOC triangle the angle BOC = 90, the angle BCO = BCD / 2 = 60/2 = 30, the angle CBO = ABC / 2 = 120/2 = 60.

Answer: The angles of the triangle BОС are equal to 30, 60, 90.