1. **State the problem:** We have a triangle ABC with angles $A=47^\circ$, $C=66^\circ$, and the perimeter is 75 cm. We need to find side $b$ to the nearest tenth. 2. **Find angle B:** Since the sum of angles in a triangle is $180^\circ$, $$ B = 180^\circ - A - C = 180^\circ - 47^\circ - 66^\circ = 67^\circ. $$ 3. **Use the Law of Sines:** Let sides opposite to angles $A, B, C$ be $a, b, c$ respectively. Then, $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \quad \text{(some constant)}. $$ 4. **Express sides in terms of $k$:** $$ a = k \sin 47^\circ, \quad b = k \sin 67^\circ, \quad c = k \sin 66^\circ. $$ 5. **Perimeter equation:** $$ a + b + c = 75 \implies k(\sin 47^\circ + \sin 67^\circ + \sin 66^\circ) = 75. $$ 6. **Calculate sums of sines:** Using approximate values, $$ \sin 47^\circ \approx 0.7314, \quad \sin 67^\circ \approx 0.9205, \quad \sin 66^\circ \approx 0.9135. $$ So, $$ k(0.7314 + 0.9205 + 0.9135) = 75 \Rightarrow k(2.5654) = 75 \Rightarrow k = \frac{75}{2.5654} \approx 29.24. $$ 7. **Find $b$:** $$ b = k \sin 67^\circ \approx 29.24 \times 0.9205 = 26.9 \text{ cm (to nearest tenth).} $$ **Final answer:** $b \approx 26.9$ cm.