In an isosceles triangle ABC with the base AC, the bisector BN of the outer angle is drawn at the vertex B.

In an isosceles triangle ABC with the base AC, the bisector BN of the outer angle is drawn at the vertex B. Determine the angle 2 if the angle 1 is 71 °.

Because BN is the bisector of the outer angle of the triangle, then the outer angle of the triangle at the vertex B is equal to:
71 * 2 = 142.
Outside angle theorem for a triangle: The outside angle is equal to the sum of two inside angles that are not adjacent to it.
∠A + ∠BCA = 142.
▲ ABC is isosceles, angles at the base are equal, so ∠BCA = 71.
The answer is ∠2 = 71 degrees