1. **State the problem:** We want to approximate the ratio $\frac{BC}{AB}$ in triangle ABC, where angle $A=55^\circ$ and angle $C=90^\circ$. 2. **Identify the relevant triangle:** Triangle ABC has a right angle at $C$ and angle $A=55^\circ$. This matches the angle configuration of Triangle 2, which has angles $35^\circ$, $55^\circ$, and $90^\circ$. 3. **Recall the trigonometric ratio:** In a right triangle, the ratio of the side opposite an angle to the hypotenuse is the sine of that angle. Here, $BC$ is opposite angle $A$, and $AB$ is the hypotenuse. So, $$\frac{BC}{AB} = \sin(55^\circ)$$ 4. **Use Triangle 2's side lengths to approximate:** Triangle 2 has sides 8.2, 10, and 5.7. The hypotenuse is 10. The side opposite $55^\circ$ is 5.7. So, $$\frac{BC}{AB} \approx \frac{5.7}{10} = 0.57$$ 5. **Check with sine value:** $\sin(55^\circ) \approx 0.819$. The ratio from Triangle 2 is 0.57, which is closer to option A (0.57) than the sine value, but since the problem asks to use one of the triangles, we trust the side ratio. **Final answer:** A) 0.57