1. **State the problem:** In triangle $\triangle ABC$, we are given that $\tan C = \frac{\sqrt{3}}{5}$ and side $AB = 110$ cm. We need to show that $BC = \frac{550}{\sqrt{3}}$ and then find the value of $b = BC$ to the nearest centimetre. 2. **Recall the tangent definition:** In a right triangle, $\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}$. 3. **Identify sides relative to angle $C$:** At angle $C$, the side opposite is $AB = 110$ cm, and the side adjacent is $BC = b$ cm. 4. **Write the tangent ratio:** $$\tan C = \frac{AB}{BC} = \frac{110}{b}$$ 5. **Substitute the given value:** $$\frac{110}{b} = \frac{\sqrt{3}}{5}$$ 6. **Solve for $b$:** $$b = \frac{110 \times 5}{\sqrt{3}} = \frac{550}{\sqrt{3}}$$ 7. **Simplify and approximate $b$:** Rationalize the denominator: $$b = \frac{550}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{550 \sqrt{3}}{3}$$ Calculate the numerical value: $$b \approx \frac{550 \times 1.732}{3} = \frac{952.6}{3} \approx 317.53$$ 8. **Final answer:** $$b \approx 318 \text{ cm (to the nearest centimetre)}$$