In the rectangle ABCD, the diagonals BD and AC are drawn, and the angle ABD = 48 degrees. Find: COD and CAD corners.

In the rectangle, the diagonals are equal and at the point of their intersection they are divided in half, then AO = BO = CO = DO. So the triangle AOB is isosceles, then the angle BAO = ABO = 48.
Since the sum of the interior angles of the triangle is 180, the angle AOB = (180 – 48 – 48) = 84.
The СOD angle and the AOB angle are the vertical angles at the intersection of the diagonals, then the СOD angle = AOB = 84.
The angle of the ВAD is straight, then the angle of the СAD = 90 – ВAO = 90 – 48 = 42.
Answer: The СOD angle is 84, the СAD angle is 42.