1. **State the problem:** Find the area of triangle ABC given angles $A=64^\circ$, $C=72^\circ$, and sides $AC=10$ units, $CB=13$ units. 2. **Identify the missing angle:** Since the sum of angles in a triangle is $180^\circ$, angle $B$ is $$B = 180^\circ - A - C = 180^\circ - 64^\circ - 72^\circ = 44^\circ.$$ 3. **Use the Law of Sines to find side $AB$:** $$\frac{AB}{\sin C} = \frac{CB}{\sin A} \implies AB = \frac{CB \cdot \sin C}{\sin A} = \frac{13 \cdot \sin 72^\circ}{\sin 64^\circ}.$$ Calculate the sines: $$\sin 72^\circ \approx 0.9511, \quad \sin 64^\circ \approx 0.8988.$$ So, $$AB = \frac{13 \times 0.9511}{0.8988} \approx \frac{12.3643}{0.8988} \approx 13.75.$$ 4. **Use the trigonometric area formula:** $$\text{Area} = \frac{1}{2} \times AC \times AB \times \sin B = \frac{1}{2} \times 10 \times 13.75 \times \sin 44^\circ.$$ Calculate $\sin 44^\circ \approx 0.6947$, so $$\text{Area} = 0.5 \times 10 \times 13.75 \times 0.6947 = 5 \times 13.75 \times 0.6947 = 67.5 \times 0.6947 \approx 46.9.$$ 5. **Check for consistency:** The area calculated is approximately 46.9 square units, which is not among the answer choices. Let's try using the given formula directly with sides $AC=10$, $CB=13$, and angle $C=72^\circ$: Using the formula $\text{Area} = \frac{1}{2} ab \sin C$ where $a=AC=10$, $b=CB=13$, and $C=72^\circ$: $$\text{Area} = \frac{1}{2} \times 10 \times 13 \times \sin 72^\circ = 65 \times 0.9511 = 61.8.$$ 6. **Final answer:** The area of triangle ABC is approximately $61.8$ square units. **Answer:** 61.8 square units.