In the rectangle ABCK, the diagonals intersect at the point O. The angle ABK is greater than the angle ABC

In the rectangle ABCK, the diagonals intersect at the point O. The angle ABK is greater than the angle ABC by 20 degrees. Find the angles of the triangle AOK.

Let the value of the angle СBК = X0, then, by condition, the angle ABK = (X + 20) 0.

Angle ABC = 90, then (X + X + 20) = 90.

2 * X = 90 – 20 = 70.

X = CBK = 70/2 = 35.

Angle СBК = AKB = 35 as criss-crossing angles at the intersection of parallel straight lines ВС and AK secant ВС.

The AOK triangle is isosceles, since OA = OK as halves of the AC and BK diagonals. Then the angle OAK = OKA = 35.

The sum of the interior angles of the triangle is 180, then the angle AOK = (180 – 35 – 35) = 110.

Answer: The angles of the AOK triangle are 35, 110, 35.