The angles formed by the side of the rhombus and its diagonals are 2 to 7. Find the corners of the rhombus.

Let the angle OAB = 2 * X0, then the angle OBA = 7 * X0.

The AOB triangle is rectangular, since the diagonals of the rhombus intersect at right angles.

The sum of the acute corners of the rectangle is 90, then 2 * X + 7 * X = 90.

X = 10.

Then the angle ОАВ = 20, the angle ОВА = 70.

The diagonals of the rhombus are the bisectors of the angles at its vertices, then the angle ABC = 2 * OBA = 2 * 70 = 140, the angle BAD = 2 * OAB = 2 * 20 = 40.

Answer: The angles of the rhombus are 140 and 40.