In triangle ABC, angle A is 50 degrees greater than angle B, and angle C is one-fifth of their sum.

Let the value of the angle ABC = X0, then, by condition, the value of the angle BAC = (50 + X) 0, and the value of the angle ACB = (X + X + 50) / 5 = (2 * X + 50) / 5.

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

180 = X + X + 50 + (2 * X + 50) / 5.

180 * 5 = 5 * X + 5 * X + 250 + 2 * X + 50.

12 * X = 600.

X = ABC = 50.

Angle BAC = 50 + 50 = 100.

Angle ACB = (2 * 50 + 50) / 5 = 30.

Answer: The angles of the triangle are 30, 50, 100.



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