The segments AC and BD are the diameters of the circle centered on O. The angle ACB is 16 °. Find the angle AOD.

Since AC and BD are the diameters of a circle centered at point O, they intersect at this point and are divided by it in half into equal segments, then:

OA = OB = OC = OD.

1. Consider △ BOC: OB = OC, so △ BOC is isosceles, then ∠OCB (aka ∠ACB) = ∠OBC = 16 ° (since these are the angles at the base of an isosceles triangle).

By the theorem on the sum of the angles of a triangle:

∠OBC + ∠BOC + ∠OCB = 180 °;

16 ° + ∠BOC + 16 ° = 180 °;

∠BOC = 180 ° – 32 °;

∠BOC = 148 °.

1. ∠BOC = ∠AOD, since they are the vertical angles formed at the intersection of two lines.

Then:

∠AOD = 148 °.

Answer: ∠AOD = 148 °.