1. **Problem statement:** In triangle ABC, given $\angle A = 40^\circ$, $\angle B = 80^\circ$, and side $b = 14.4$ feet, find $\angle C$ and side $c$. 2. **Step 1: Find $\angle C$** The sum of angles in any triangle is $180^\circ$. So, $$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 40^\circ - 80^\circ = 60^\circ.$$ 3. **Step 2: Use the Law of Sines to find side $c$** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$ We know $b = 14.4$, $\angle B = 80^\circ$, and $\angle C = 60^\circ$. So, $$\frac{c}{\sin 60^\circ} = \frac{14.4}{\sin 80^\circ}.$$ 4. **Step 3: Calculate $c$** Rearranging, $$c = \frac{14.4 \times \sin 60^\circ}{\sin 80^\circ}.$$ Using approximate values, $$\sin 60^\circ \approx 0.866, \quad \sin 80^\circ \approx 0.985.$$ So, $$c \approx \frac{14.4 \times 0.866}{0.985} \approx \frac{12.4704}{0.985} \approx 12.66.$$ 5. **Final answers:** $$\angle C = 60^\circ, \quad c \approx 12.66 \text{ feet}.$$