Consider a triangle $ABC$ with $\angle ACB=120^\circ$. Let $A’, B’, C’$ be the points of intersection of the angular bisector through $A$, $B$ and $C$ with the opposite side, respectively.
Determine $\angle A’C’B’$.
1. **Identify the given information and the goal:**
   - We are given a triangle \(ABC\) with \(\angle ACB = 120^\circ\).
   - Points \(A'\), \(B'\), and \(C'\) are the points of intersection of the angle bisectors of \(\angle A\), \(\angle B\), and \(\angle C\) with the opposite sides, respectively.
   - We need to determine \(\angle A'C'B'\).

2. **Understand the properties of the angle bisectors:**
   - The angle bisector of \(\angle A\) intersects \(BC\) at \(A'\).
   - The angle bisector of \(\angle B\) intersects \(AC\) at \(B'\).
   - The angle bisector of \(\angle C\) intersects \(AB\) at \(C'\).

3. **Use the given angle \(\angle ACB = 120^\circ\):**
   - Since \(\angle ACB = 120^\circ\), the external angle at \(C\) is \(180^\circ - 120^\circ = 60^\circ\).

4. **Consider the properties of the angle bisectors:**
   - The angle bisector of \(\angle ACB\) will split \(\angle ACB\) into two equal angles of \(60^\circ / 2 = 30^\circ\).
   - Therefore, \(\angle A'CB = 30^\circ\) and \(\angle B'CA = 30^\circ\).

5. **Analyze the triangle \(A'C'B'\):**
   - Since \(A'\) and \(B'\) are the points where the angle bisectors intersect the opposite sides, \(A'C'\) and \(B'C'\) are angle bisectors of \(\angle ACB\).
   - The angle bisectors of \(\angle ACB\) will meet at the incenter of \(\triangle ACB\).

6. **Determine \(\angle A'C'B'\):**
   - Since \(A'C'\) and \(B'C'\) are angle bisectors, they will form right angles with the sides of the triangle at the points where they intersect.
   - Therefore, \(\angle A'C'B'\) is a right angle.

\[
\angle A'C'B' = 90^\circ
\]

\(\blacksquare\)

The final answer is \( \boxed{ \angle A'C'B' = 90^\circ } \)