One of the outer corners of an isosceles triangle is 70 degrees. Find the corners of the triangle.

Let △ ABC be given, in which AB = BC, then AC is the base of this triangle, therefore ∠A = ∠C as angles at the base of an isosceles triangle.
1. Let ∠BAD = 70 ° be the outer angle at the vertex A, then ∠A = 180 ° – ∠BAD = 180 ° – 70 ° = 110 °.
If ∠A = ∠C, then ∠C = 110 °.
Two angles at the base of an isosceles triangle cannot be obtuse, since a triangle cannot have two obtuse angles.
2. Let ∠CBD = 70 ° be the outer angle at vertex B, then ∠B = 180 ° – ∠CBD = 180 ° – 70 ° = 110 °.
Since ∠A = ∠C, we denote them as x. By the theorem on the sum of the angles of a triangle:
∠A + ∠B + ∠C = 180 °;
x + 110 ° + x = 180 °;
2 * x = 180 ° – 110 °;
2 * x = 70 °;
x = 70 ° / 2;
x = 35 °.
Thus, ∠A = ∠C = x = 35 °.
Answer: ∠A = 35 °, ∠B = 110 °, ∠C = 35 °.