In the rectangle, the diagonals AC and BD meet at point O. Find the angle ACB if the angle AOD = 110.

Angle BOC = AOD = 110 as the vertical angles at the intersection of the diagonals AC and BD.

In a rectangle, the diagonals are equal and at the point of their intersection they are divided in half, then BO = OС.

Since BО = OC, then the triangle BОС is isosceles, then the angle ОВС = OCB.

The sum of the inner angles of the triangle is 180, then the OCB angle = (180 – 110) / 2 = 70/2 = 35.

Point O belongs to the diagonal AC, then the angle ACB = OCB = 35.

Answer: Angle ACB is 35