1. **State the problem:** We need to find the area of triangle ABC where side AC = 24 cm, side AB = 25 cm, and the angle at vertex A is 75°. 2. **Formula used:** The area of a triangle when two sides and the included angle are known is given by: $$\text{Area} = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are the sides enclosing the angle $C$. 3. **Identify sides and angle:** Here, $a = 24$ cm, $b = 25$ cm, and $C = 75^\circ$. 4. **Calculate the sine of the angle:** $$\sin(75^\circ) \approx 0.9659$$ 5. **Calculate the area:** $$\text{Area} = \frac{1}{2} \times 24 \times 25 \times 0.9659$$ 6. **Simplify step-by-step:** $$= 12 \times 25 \times 0.9659$$ $$= 300 \times 0.9659$$ $$= 289.77$$ 7. **Final answer rounded to 1 decimal place:** $$\boxed{289.8\text{ cm}^2}$$