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

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

In a convex quadrilateral OKCH, the sum of its interior angles is 360.

The angle HOK, by condition, is equal to 100, the angle OKC = OНС = 90, since AK and BH are the heights of the ABC triangle.

Then the angle KCH = (360 – HOK – OKС – OНС) = (360 – 100 – 90 – 90) = 80.

Since the triangle ABC is isosceles, the angle BAC = BCA = 80.

The sum of the inner angles of the triangle is 180, then the angle ABC = (180 – BAC – BCA) = (180 – 80 – 80) = 20.

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