The bisector of an isosceles triangle drawn from the apex at the base and forms

The bisector of an isosceles triangle drawn from the apex at the base and forms angles with the opposite side, one of which is 60 degrees. Find the angles of this triangle.

Since the triangle ABC is isosceles, its angles at the base of the AC are equal.

Angle BAC = BCA.

Let the value of the angle BCA = 2 * X0, then the angle BAC = 2 * X0.

Since AM is the bisector of the angle BAC, the angle CAM = BAC / 2 = 2 * X / 2 = X0.

The angles AMC and AMB are adjacent, the sum of which is 180, then the angle AMC = (180 – 60) = 120.

The sum of the interior angles of the AFM triangle is 180.

Then 180 = X + 2 * X + 120.

3 * X = 60.

X = 20.

Then the angle BAC = BCA = 2 * 20 = 40.

Angle ABC = (180 – 40 – 40) = 100.

Answer: The angles of the triangle are 100, 40, 40.