In an isosceles triangle ABC with base AC from the vertices A and B, heights are drawn which

In an isosceles triangle ABC with base AC from the vertices A and B, heights are drawn which, when they intersect, form an angle of 100. Find the angles of the triangle.

Angles AOB and BOK are adjacent angles, the sum of which is 180, then the angle BOK = (180 – 100) = 80.

In a right-angled triangle BOK, the angle OBK = (180 – 90 – 80) = 10.

The height BH of the triangle ABC is also its bisector, then the angle ABC = 2 * OBK = 2 * 10 = 20.

Angle ABC = BCA, as angles at the base of an isosceles triangle.

Then the angle BAC = BCA = (180 – 20) / 2 = 160/2 = 80.

Answer: The angles of the triangle ABC are equal to 80, 80, 20.