Given: rhombus ABCD, diagonals intersect at point O, angle AOB = 80 degrees. Find the angles of triangle AOB.

Let us have a rhombus ABCD, the point of intersection of the diagonals of which is O. Let the angle BAD = 80 degrees. Consider the triangle AOB. Since the diagonals of the rhombus intersect at right angles, we have that the BOA angle is 90 degrees. From the property of the diagonals of a rhombus, we know that they are the bisectors of its angles. Then the angle BAD will be equal to two angles of the BAO. From here we get that VAO = 80/2 = 40 (degrees).
From the fact that the sum of the angles of the triangle is 180 degrees, we have that ABO = 180 – (BOA + BAO)
Then ABO = 180 – (40 + 90) = 50 (degrees). Thus, we get that the angles of our triangle are 50, 40, 90 degrees.
Answer: 50, 40, 90 degrees.