The heights of the isosceles triangle, drawn from the vertices at the base, at the intersection

The heights of the isosceles triangle, drawn from the vertices at the base, at the intersection form an angle of 140 degrees. Find the corner opposite the base.

The value of the angle AOC, according to the condition, wound 140, then the angle C1OA1 = AOC = 140, as the vertical angles at the intersection of straight lines CC1 and AA1.

Consider a quadrangle OC1BA1, in which the angles CC1B = AA1B = 90, and the angle A1OC1 = 1400, then the angle A1BC1 = ABC = 360 – 140 – 90 – 90 = 40.

Since the triangle is isosceles, the angle BAC = BCA = (180 – 40) / 2 = 70.

Answer: The angle at the base is 70.