In a triangle ABC = 40º, the outer angle at the vertex B is 70º. Find the rest of the interior corners of the triangle.

Apparently 40 ° is the inner angle at the top of C.
The problem can be solved in different ways, consider these options.
Option 1.
Knowing the outer angle at the vertex B, we find the inner angle B:
∠ B = 180 ° – 70 ° = 110 °.
Two angles of the triangle are known, we find the third – angle A.
∠ А = 180 ° – ∠ В – ∠ С = 180 ° – 110 ° – 40 ° = 30 °.
Option 2.
We find the inner angle B in the same way as in the first case:
∠ B = 180 ° – 70 ° = 110 °.
Let’s use one of the properties of the outer corner of the triangle. Its degree measure is equal to the sum of two interior angles that are not adjacent to it. In our case, we get:
∠ In ext. = ∠ А + ∠ С → ∠ А = ∠ В ext. – ∠ С = 70 ° – 40 ° = 30 °.
Answer: angle B = 110 °, angle A = 30 °.