The bisector AL is drawn in the triangle ABC, the angle ALC is 169 °, the angle ABC is 160 °. 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).
160 ° + ∠ACB + 2x = 180 °
∠ACB + 2x = 20 °
x = (20 ° -∠ACB) / 2
Consider the triangle ALC:
∠ALC + ∠ACB + ∠LAC = 180 ° (by the theorem on the sum of the angles of a triangle).
169 ° + ∠ACB + x = 180 °
∠ACB + x = 11 °
Substitute the x value obtained earlier:
∠ACB + (20 ° -∠ACB) / 2 = 11 ° | * 2
2∠ACB + 20 ° -∠ACB = 22 °
∠ACB = 22 ° -20 ° = 2 °
Answer: 2