In an isosceles triangle, the outer angle at the base is 140 degrees. Find the angles of this triangle.

Let the triangle ABC be given, the angle DAV is the outer angle at the base. Angle DAB = 140 degrees according to the assignment. The corners DAB and BAC are adjacent. The sum of adjacent angles is 180 degrees.
Let’s find the angle BAC 180-angle DAB. BAC = 180-140 = 40 degrees.
In an isosceles triangle, the angles at the base are equal. Accordingly, the angle BAC = the angle of the BCA. ICA angle = 40 degrees.
The sum of the degrees of the angles of the triangle is 180 degrees.
We get that the angle ABC = 180-angle BCA-angle BAC. angle ABC = 180-40-40 = 100 degrees
Answer: The angles of the triangle are 100, 40 and 40 degrees.