AC and BD are diameters of a circle centered on O. Center angle AOD is 118 °. Find the inscribed angle ACB.

The AOD angle is equal to the BOC angle as the vertical angles at the intersection of the AC and BD diameters, then the BOC angle = 118.

The BОС triangle is isosceles, since its sides ОВ and ОВ are the radii of the circle, then the angle СBО = ОСB = (180 – BОС) / 2 = (180 – 118) / 2 = 31.

Angle ACB = OCB = 31, since point O lies on the diagonal AC

Answer: Angle AOD is 31.

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