In an isosceles triangle ABC with a base BC, the outer angle at apex A is 130 degrees.

In an isosceles triangle ABC with a base BC, the outer angle at apex A is 130 degrees. Find the outside angle at vertex B.

An isosceles triangle is a triangle in which two sides (sides) have the same length, and the third side is its base. In such a triangle, the angles at the base are the same.

Let us find the value of one internal angle at the base of the BC, taking into account that the value of the external angle is equal to the sum of two internal angles that will not be adjacent to it, that is

∠В + ∠С = 130 °, and since ∠В = ∠С, then ∠В = ∠С = 130 °: 2 = 65 °.

Since the sum of the inner and outer angles at one vertex is equal to 180 °, we find the value of the outer angle at the vertex B (as well as C):

180 – 65 = 115 °.

Answer: 115 °.