To determine which plane reaches a greater horizontal distance from the airport at an altitude of 10,000 feet, we can use trigonometry. We can use the tangent function, which relates the angle of elevation to the opposite side (altitude) and the adjacent side (horizontal distance).
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For Plane A with a 13° angle: \[ \tan(13°) = \frac{\text{Altitude}}{\text{Horizontal Distance A}} \] Rearranging gives: \[ \text{Horizontal Distance A} = \frac{\text{Altitude}}{\tan(13°)} = \frac{10,000 \text{ feet}}{\tan(13°)} \]
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For Plane B with a 9° angle: \[ \tan(9°) = \frac{\text{Altitude}}{\text{Horizontal Distance B}} \] Rearranging gives: \[ \text{Horizontal Distance B} = \frac{\text{Altitude}}{\tan(9°)} = \frac{10,000 \text{ feet}}{\tan(9°)} \]
Now calculating each:
For Plane A:
\[ \tan(13°) \approx 0.2309 \] \[ \text{Horizontal Distance A} = \frac{10,000}{0.2309} \approx 43,309 \text{ feet} \]
For Plane B:
\[ \tan(9°) \approx 0.1584 \] \[ \text{Horizontal Distance B} = \frac{10,000}{0.1584} \approx 63,158 \text{ feet} \]
Summary of Results:
- Plane A: 43,309 feet
- Plane B: 63,158 feet
Thus, Plane B reaches a greater horizontal distance from the airport when it reaches an altitude of 10,000 feet.
Final Answer:
Plane B because it was 63,138 feet away (rounded from 63,158 feet).
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