The bisector AL is drawn in the triangle ABC, the angle ALC is 148 °, the angle ABC is 132 °. Find the angle of the ACB.

Let ∠BAL = x
Then, ∠LAC also = x (since AL is a bisector).
Consider the triangle ABC:
∠ABC + ∠ACB + ∠CAB = 180 ° (by the theorem on the sum of the angles of a triangle).
132 ° + ∠ACB + 2x = 180 °
∠ACB + 2x = 48 °
x = (48 ° -∠ACB) / 2
Consider the triangle ALC:
∠ALC + ∠ACB + ∠LAC = 180 ° (by the theorem on the sum of the angles of a triangle).
148 ° + ∠ACB + x = 180 °
∠ACB + x = 32 °
Substitute the x value obtained earlier:
∠ACB + (48 ° -∠ACB) / 2 = 32 ° | * 2
2∠ACB + 48 ° -∠ACB = 64 °
∠ACB = 64 ° -48 ° = 16 °
Answer: 16