In the triangle ABC, AC = BC. The external angle at vertex B is 155 °. Find the angle C

∠CBA – is adjacent to the outer corner, therefore 180 ° = ∠CBA + 155 °
∠CBA = 180 ° -155 ° = 25 °
Since AC = BC, the triangle ABC is isosceles.
So ∠CBA = ∠CAB = 25 ° (by the property of an isosceles triangle)
By the theorem on the sum of the angles of a triangle:
180 ° = ∠CBA + ∠CAB + ∠C
180 ° = 25 ° + 25 ° + ∠C
∠C = 130 °
Answer: 130