1. **State the problem:** Find the area of triangle ABC where sides AC = 17 cm, AB = 16 cm, and the angle at vertex C is 65°. 2. **Formula used:** The area of a triangle given two sides and the included angle is $$\text{Area} = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are the sides enclosing angle $C$. 3. **Identify sides and angle:** Here, $a = 17$ cm, $b = 16$ cm, and $C = 65^\circ$. 4. **Calculate the area:** $$\text{Area} = \frac{1}{2} \times 17 \times 16 \times \sin(65^\circ)$$ 5. **Simplify multiplication:** $$= \frac{1}{2} \times 272 \times \sin(65^\circ)$$ 6. **Calculate half of 272:** $$= 136 \times \sin(65^\circ)$$ 7. **Evaluate $\sin(65^\circ)$:** $$\sin(65^\circ) \approx 0.9063$$ 8. **Final area calculation:** $$\text{Area} \approx 136 \times 0.9063 = 123.26$$ **Answer:** The area of triangle ABC is approximately $123.26$ cm$^2$.