To find the ratio of the side opposite ∠A to the hypotenuse of triangle ABC, we first identify the sides involved. In triangle ABC, the side opposite ∠A is side BC, which has a length of 8, and the hypotenuse is side CA, which has a length of 10. The ratio can be calculated as \( \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{CA} = \frac{8}{10} = \frac{4}{5} \).
For the second triangle, with corresponding side lengths of 32, 40, and 24, we apply the same ratio to find the length of the side opposite ∠A', which is given by \( \frac{4}{5} \) of the hypotenuse. The hypotenuse of the second triangle is 40, so the side opposite ∠A' would be \( \frac{4}{5} \times 40 = 32 \). Thus, point A' is located at the vertex opposite side B'C' in the second triangle, with the corresponding side lengths ensuring that angle A' maintains the same sine ratio as angle A in triangle ABC.