In a circle centered on O AC and BD diameters, the AOD center angle is 130 degrees, find the inscribed angle ACB.

univerkov | September 26, 2021 | education

The triangle AOD is isosceles, since OA and OD are the radii of a circle, then the angle OAD = ODA.

Then the angle OAD = ODA = (180 – AOD) / 2 = (180 – 130) / 2 = 50/2 = 250.

The inscribed angles BDA and ACB rest on one arc AB, then the angle ACB = BDA = 25.

Second way.

Angle BOC = AOD = 130 as vertical angles. The triangle BOС is isosceles, then the angle OBC = OCB = ACB = (180 – BOС) / 2 = (180 – 130) / 2 = 25.

Answer: The inscribed angle ACB is 25.

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