In a circle with center O AC and BD are diameters. The center angle of the AOD is 68 degrees.

In a circle with center O AC and BD are diameters. The center angle of the AOD is 68 degrees. Find the inscribed corner ACB.

Since AC and BD are the diameters of the circle, and point O is the point of their intersection, then the angles of AOD and AOB are vertical angles, the sum of which is 180, then the angle AOB = (180 – AOD) = (180 – 68) = 112.

The central angle AOB rests on the arc AB, the value of which is equal to the degree measure of the angle AOB.

Arc AB = 112. The inscribed angle ACB also rests on the arc AB, then the angle ACB is equal to half the degree measure of the arc AB.

Angle ACB = 112/2 = 56.

Answer: The inscribed angle ACB is 56.