1. **State the problem:** Given triangle ABC with angles $A=41^\circ$, $B=82^\circ$, and side $a=56$, find the missing angle $C$ and sides $b$ and $c$. 2. **Use the triangle angle sum rule:** The sum of angles in a triangle is $180^\circ$. $$ C = 180^\circ - A - B = 180^\circ - 41^\circ - 82^\circ = 57^\circ $$ 3. **Use the Law of Sines:** $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$ 4. **Calculate side $b$:** $$ b = a \times \frac{\sin B}{\sin A} = 56 \times \frac{\sin 82^\circ}{\sin 41^\circ} $$ Calculate the sines: $\sin 82^\circ \approx 0.9903$, $\sin 41^\circ \approx 0.6561$ $$ b \approx 56 \times \frac{0.9903}{0.6561} \approx 56 \times 1.509 = 84.5 $$ 5. **Calculate side $c$:** $$ c = a \times \frac{\sin C}{\sin A} = 56 \times \frac{\sin 57^\circ}{\sin 41^\circ} $$ Calculate the sines: $\sin 57^\circ \approx 0.8387$ $$ c \approx 56 \times \frac{0.8387}{0.6561} \approx 56 \times 1.279 = 71.6 $$ **Final answers rounded to the nearest tenth:** $$ C = 57^\circ, \quad b = 84.5, \quad c = 71.6 $$