The bisector AL is drawn in the triangle ABC, the angle ALC is 121 degrees, the angle ABC is 101 degrees, find the angle ACB.

1.∠ALB and ∠ALC are adjacent, then:
∠ALB + ∠ALC = 180 °.
Thus:
∠ALB + 121 ° = 180 °;
∠ALB = 180 ° – 121 °;
∠ALB = 59 °.
2. Consider △ ABL:
∠LAB + ∠ABL (aka ∠ABC) + ∠ALB = 180 ° (by the theorem on the sum of the catch of a triangle).
Thus:
∠LAB + 101 ° + 59 ° = 180 °;
∠LAB = 180 ° – 160 °;
∠LAB = 20 °.
3. Since AL is a bisector, then:
∠LAB = ∠LAC.
Thus:
∠LAC = 20 °.
4. Consider △ ALC:
∠LAC + ∠ALC + ∠ACL = 180 ° (by the theorem on the sum of the catch of a triangle) /
Thus:
20 ° + 121 ° + ∠ACL = 180 °;
∠ACL = 180 ° – 141 °;
∠ACL = 39 °.
∠ACL = ∠ACB = 39 °.
Answer: ∠ACB = 39 °.