In triangle ABC, angle A is 20 ° greater than angle C, angle C is 6 times less than angle B. Find the angle of the triangle ABC.

Let the value of the angle ACB be equal to X0, then, by condition, the value of the angle BAC = (X + 20) 0, and the value of the angle ABC = 6 * X0.

The sum of the interior angles of the triangle is 1800, then:

180 = (ABC + BAC + ACB) = (6 * X + (X + 20) + X).

8 * X = 180 – 20 = 160.

X = 160/8 = 20.

Angle ACB = 20.

Angle BAC = 20 + 20 = 40.

Angle ABC = 6 * 20 = 120.

Answer: The angles of the triangle ABC are equal to 20, 40, 120.