1. **Problem statement:** We have triangle A-C-B with vertex C at the top, base AB horizontal, side AC labeled $b$, side CB labeled $a=8$, and base AB labeled $c$. Given angles $\angle A = 47^\circ$ and $\angle C = 88^\circ$, find the length of side $b = AC$. 2. **Step 1: Find the missing angle $\angle B$.** Since the sum of angles in a triangle is $180^\circ$, $$\angle B = 180^\circ - \angle A - \angle C = 180^\circ - 47^\circ - 88^\circ = 45^\circ.$$ 3. **Step 2: Use the Law of Sines.** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$ We know $a=8$, $\angle A=47^\circ$, $\angle B=45^\circ$, and want to find $b$. 4. **Step 3: Solve for $b$.** $$b = a \times \frac{\sin B}{\sin A} = 8 \times \frac{\sin 45^\circ}{\sin 47^\circ}.$$ 5. **Step 4: Calculate the sines and simplify.** $$\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071,$$ $$\sin 47^\circ \approx 0.7314.$$ 6. **Step 5: Substitute values and compute.** $$b = 8 \times \frac{0.7071}{0.7314} = 8 \times 0.9667 = 7.7336.$$ **Final answer:** $$\boxed{b \approx 7.73}.$$