The inner angles of the triangle are proportional to the numbers 3 5 1 find the inner and outer corners of the triangle.

Let the value of the angle CAB = X0, then, by condition, the value of the angle ACB = 3 * X0, and the angle ABC = 5 * X0.

The sum of the inner angles of the triangle is 1800, then (X + 3 * X + 5 * X) = 180.

9 * X = 180.

X = 180/9 = 20.

Then the angle CAB = 1 * 20 = 20, angle ACB = 3 * 20 = 60, angle ABC = 5 * 20 = 100.

Let’s define the outer corners of the triangle ABC.

Angle ACC1 = (180 – ASB) = (180 – 60) = 120, angle BAA1 = (180 – CAB) = (180 – 20) = 160, angle ABB1 = (180 – ABC) = (180 – 100) = 80 …

Answer: The angles of the triangle are 20, 60, 100, the outer corners of the triangle are 80, 120, 160.