In a circle centered at point O, there are segments AC and BD diameters. Angle AOD = 148. Find the angle ACB.

Angle BOC = AOD = 148 as vertical angles at the intersection of segments AC and OD.

The ВOС triangle is isosceles, since the segments AC and BD are diameters, and then OB = OС = R.

In an isosceles triangle ВOС, the angle OСВ = OBC = (180 – 140) / 2 = 40/2 = 20.

Point O lies on the segment AC, then the angle OCB = CB = 20.

Answer: The ACB angle is 20.

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