ABCD is a rectangle, O is the point of intersection of its diagonals. The AOB angle is 48 °. Find the ADB angle.

<AOB = COD = 48 °.

BO = AO (the diagonals of the rectangle are equal and at the point of intersection they divide each other in half).

ΔAOB – isosceles.

<ABO = <BAO;

<ABO + <BAO + <AOB = 180 °;

<BAO + <BAO + <AOB = 180 °;

2 * <BAO + 48 ° = 180 °;

<BAO = (180 ° – 48 °) / 2 = 66 °.

<OAD = 90 ° – <BAO = 90 ° – 66 ° = 24 °;

Triangle AOB – isosceles:

AO = OD;

<ADO = <OAD = 24 °.

Answer: 24 °.

A simple solution that requires additional construction is to draw the horizontal axis of symmetry through the intersection of the diagonals. The axis will divide the given angle in half: 48 ° / 2 = 24 °. The desired angle and half of the given angle will be one-sided matching at the intersection of parallel lines.