1. **State the problem:** We are given a triangle with angles $A=40^\circ$, $B=45^\circ$, and side $c=28$ ft opposite angle $C$. We need to find angle $C$, side $a$ opposite angle $A$, and side $b$ opposite angle $B$ using the Law of Sines. 2. **Find angle $C$:** The sum of angles in a triangle is $180^\circ$. $$C = 180^\circ - A - B = 180^\circ - 40^\circ - 45^\circ = 95^\circ$$ 3. **Law of Sines formula:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 4. **Calculate side $a$:** $$a = c \times \frac{\sin A}{\sin C} = 28 \times \frac{\sin 40^\circ}{\sin 95^\circ}$$ Calculate intermediate values: $$\sin 40^\circ \approx 0.6428, \quad \sin 95^\circ \approx 0.9962$$ $$a = 28 \times \frac{0.6428}{0.9962} = 28 \times 0.6453 = 18.049$$ 5. **Calculate side $b$:** $$b = c \times \frac{\sin B}{\sin C} = 28 \times \frac{\sin 45^\circ}{\sin 95^\circ}$$ Calculate intermediate values: $$\sin 45^\circ \approx 0.7071$$ $$b = 28 \times \frac{0.7071}{0.9962} = 28 \times 0.7097 = 19.871$$ 6. **Final answers rounded to three decimal places:** $$C = 95^\circ$$ $$a = 18.049 \text{ ft}$$ $$b = 19.871 \text{ ft}$$