1. Stating the problem: We need to find an approximate value for $\frac{\cos 2^\circ}{\sin 1^\circ}$ given that $1^\circ=0.018$ radians and $(0.018)^2=0.000324$. The answer should be correct to 2 decimal places. 2. Convert the angles to radians: $2^\circ = 2 \times 0.018 = 0.036$ radians and $1^\circ = 0.018$ radians. 3. Use approximations for cosine and sine for small angles using Taylor expansions: - $\cos x \approx 1 - \frac{x^2}{2}$ - $\sin x \approx x$ 4. Calculate approximate value for $\cos 2^\circ$: $$\cos 0.036 \approx 1 - \frac{(0.036)^2}{2} = 1 - \frac{0.001296}{2} = 1 - 0.000648 = 0.999352$$ 5. Calculate approximate value for $\sin 1^\circ$: $$\sin 0.018 \approx 0.018$$ 6. Calculate the ratio: $$\frac{\cos 2^\circ}{\sin 1^\circ} \approx \frac{0.999352}{0.018} = 55.52$$ 7. Final answer: $$\boxed{55.52}$$