In a circle with center O AC and BD are diameters. The center angle of the AOD is 112 degrees.

In a circle with center O AC and BD are diameters. The center angle of the AOD is 112 degrees. Find the inscribed angle ACB.

Angle BOC = AOD = 112 as the vertical angles at the intersection of the diagonals AC and BD.

The ВOС triangle is isosceles, since OB = OС = R.

Then the angle OBC = OCВ = (180 – ВOС) / 2 = (180 – 112) / 2 = 34.

Angle ACB = OCB = 34.

Answer: The ACB angle is 340.



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