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

October 12, 2021 | education

| 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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