Find the interior angle BCA.

Angles on a straight line sum to $180^\circ$.

$$\angle BCA + 125^\circ = 180^\circ$$ $$\angle BCA = 180^\circ - 125^\circ$$ $$\angle BCA = 55^\circ$$

The markings on sides AB and BC indicate that $AB = BC$. Therefore, $\triangle ABC$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal.

$$\angle BAC = \angle BCA$$ $$\angle BAC = 55^\circ$$

The sum of angles in a triangle is $180^\circ$.

$$\angle ABC + \angle BAC + \angle BCA = 180^\circ$$ $$\angle ABC + 55^\circ + 55^\circ = 180^\circ$$ $$\angle ABC + 110^\circ = 180^\circ$$ $$\angle ABC = 180^\circ - 110^\circ$$ $$\angle ABC = 70^\circ$$