The bisector AL is drawn in triangle ABC, angle ALC is 112 °, angle ABC is 106 °. Find the ACB angle.

Angle АLC is the outer angle of triangle ABL, then its value is equal to the sum of two inner, not adjacent angles.

Angle ALC = BAL + ABL, then angle BAL = ALC – ABL = 112 – 106 = 6.

Since, by condition, AL is the bisector of angle A, then the angle CAL = BAL = 6.

In triangle ACL, angle ACL = 180 – ALC – CAL = 180 – 112 – 6 = 62.

Angle ACB = ACL = 62.

Answer: Angle ABC is 62.