In triangle ABC, we have:
- Side \(AB = 9\) (opposite angle A)
- Side \(BC = 12\)
- Side \(AC = 15\) (hypotenuse)
To find the ratio of the side opposite angle A (which is \(AB\)) to the hypotenuse (which is \(AC\)), we can calculate the following:
\[ \text{Ratio} = \frac{AB}{AC} = \frac{9}{15} = \frac{3}{5} = 0.6 \]
Now, let's analyze triangle A'B'C' with side lengths:
- \(A'B' = 30\) (opposite angle A')
- \(B'C' = 24\)
- \(A'C' = 18\) (hypotenuse)
Since this triangle is also a right triangle, we can check which side corresponds to the angle A by using the ratio we derived earlier (0.6).
Given that the ratio is \(0.6\), we look to see where it applies in triangle A'B'C'. Since the original triangle ABC had \(AB\) as the side opposite to angle A and its hypotenuse \(AC\), we look for the side opposite angle A' in triangle A'B'C' that maintains the same ratio of opposite side to hypotenuse.
Calculating for triangle A'B'C':
- Hypotenuse \(A'C' = 18\)
- Side opposite angle A' is \(A'B' = 30\)
To find the relevant ratio in triangle A'B'C': \[ \text{Ratio} = \frac{30}{18} = \frac{5}{3} \approx 1.67 \]
This is not the same as the ratio \(0.6\). Therefore, we consider the other side:
- Side \(B'C' = 24\)
Calculating the ratio with \(24\) as the opposite side: \[ \text{Ratio} = \frac{24}{18} = \frac{4}{3} \approx 1.33 \]
Both of those options do not yield a ratio of \(0.6\). We need to relate with what was given for \(A'\).
Given the hint in the question, the opposite side in triangle A'B'C' should fit our original calculations. Then it turns out the triangle stated with \(18\) (the smallest side corresponding to the original hypotenuse):
Thus regarding the last choice \( \text{The ratio of the opposite side to the hypotenuse is } 0.6\text{, and point } A' \text{ is opposite the side that has length } 18.\).
Conclusively, the correct answer is: The ratio of the opposite side to the hypotenuse is 0.6, and point A′ is opposite the side that has length 18.
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