Points ABC lie on a circle with center O, angle AOB = 80 degrees, semicircle AC: BC = 2: 3. Find triangle ABC.

The inscribed angle AOC = 80 and rests on the arc AB, then the degree measure of the arc AB = 80.

The sum of the degree measures of the arcs AC and BC is equal to: AC + BC = (360 – AB) = (360 – 80) = 280.

Let the degree measure of the arc AC = 2 * X0, then the arc BC = 3 * X0.

2 * X + 3 * X = 280.

5 * X = 280.

X = 280/5 = 56.

Then the arc AC = 2 * 56 = 112, arc BC = 3 * 56 = 168.

The inscribed angle is equal to half of the degree measure of the arc on which it rests, then the angle ABC = 112/2 = 56, the angle BAC = 168/2 = 84, the angle ACB = 80/2 = 40.

Answer: The angles of the triangle ABC are 40, 56, 84.