The bisector AL is drawn in triangle ABC, the angle ALC is 152 °, and the angle ABC is 137 °. 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).
137 ° + ∠ACB + 2x = 180 °
∠ACB + 2x = 43 °
x = (43 ° -∠ACB) / 2
Consider the triangle ALC:
∠ALC + ∠ACB + ∠LAC = 180 ° (by the theorem on the sum of the angles of a triangle).
152 ° + ∠ACB + x = 180 °
∠ACB + x = 28 °
Substitute the x value obtained earlier:
∠ACB + (43 ° -∠ACB) / 2 = 28 ° | * 2
2∠ACB + 43 ° -∠ACB = 56 °
∠ACB = 56 ° -43 ° = 13 °
Answer: 13

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