1. **State the problem:** We are given a triangle ABC with sides AB = 8 cm, BC = 10 cm, and area = 24 cm². We need to find the angle ACB (angle at vertex C). 2. **Recall the formula for the area of a triangle:** $$\text{Area} = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are two sides enclosing angle $C$. 3. **Identify the sides enclosing angle ACB:** Angle ACB is at vertex C, so the sides enclosing it are AC and BC. 4. **Find length AC:** Using the Pythagorean theorem approximation from the user's note: $$AB^2 + BC^2 = 8^2 + 10^2 = 64 + 100 = 164$$ $$AC = \sqrt{164} \approx 12.81$$ 5. **Use the area formula with sides AC and BC:** $$24 = \frac{1}{2} \times AC \times BC \times \sin(\angle ACB)$$ Substitute known values: $$24 = \frac{1}{2} \times 12.81 \times 10 \times \sin(\angle ACB)$$ 6. **Simplify and solve for $\sin(\angle ACB)$:** $$24 = 64.05 \times \sin(\angle ACB)$$ $$\sin(\angle ACB) = \frac{24}{64.05}$$ $$\sin(\angle ACB) \approx 0.3747$$ 7. **Find the angle ACB:** $$\angle ACB = \arcsin(0.3747) \approx 22.0^\circ$$ **Final answer:** The angle ACB is approximately $22.0^\circ$.