# Internal corners of triangle ABC || 2,5,8 a) find the angles of the triangle ABC b) Find the outer corners of the triangle ABC.

Let the value of the angle CAB = 2 * X0, then, by condition, the value of the angle ACB = 5 * X0, and the angle ABC = 8 * X0.

The sum of the interior angles of the triangle is 180, then (2 * X + 5 * X + 8 * X) = 180.

15 * X = 180.

X = 180/15 = 120.

Then the angle CAB = 2 * 12 = 240, angle ACB = 5 * 12 = 60, angle ABC = 8 * 12 = 96.

Determine the outer corners of the triangle ABC.

Angle ACC1 = (180 – ACB) = (180 – 60) = 120, angle BAA1 = (180 – CAB) = (180 – 24) = 156, angle ABB1 = (180 – ABC) = (180 – 96) = 84 …

Answer: The angles of the triangle are 24, 60, 96, the outer corners of the triangle are 84, 120, 156.

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