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Plane A takes off at a 13° angle from the runway, and plane B takes off at a 9° angle from the runway. Which plane reaches a greater horizontal distance from the airport when the plane reaches an altitude of 10,000 feet? Round the solutions to the nearest whole number. Plane A because it was 43,315 feet away Plane A because it was 63,925 feet away Plane B because it was 44,454 feet away Plane B because it was 63,138 feet away

User ZachRabbit
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2 Answers

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Final answer:

Using trigonometry and the tangent function, we determine that Plane B, which takes off at a lower 9° angle, reaches a greater horizontal distance from the airport when it reaches an altitude of 10,000 feet.

Step-by-step explanation:

To determine which plane reaches a greater horizontal distance from the airport when reaching an altitude of 10,000 feet, we can use trigonometry. For each plane, the altitude forms the opposite side of a right-angled triangle, and the angle of ascent is given. We can find the horizontal distance (the adjacent side of the triangle) by using the tangent function, where tan(θ) = opposite/adjacent, which yields adjacent = opposite/tan(θ). We can then calculate the distances for both planes.

For Plane A, with a 13° angle:

  • adjacent = 10,000 / tan(13°)

For Plane B, with a 9° angle:

  • adjacent = 10,000 / tan(9°)

Without calculating these values exactly, we can already tell that Plane A, with the steeper angle, will have a shorter horizontal distance compared to Plane B, because the tangent function is an increasing function. This means that as the angle decreases, the tangent value decreases, the denominator decreases and therefore the adjacent side (the horizontal distance) increases.

Therefore, Plane B reaches a greater horizontal distance from the airport when it reaches an altitude of 10,000 feet.

User PrettyInPink
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0 votes

Answer:

Set your calculator to Degree mode.

The diagrams are omitted--please sketch a diagram for each plane to confirm my answer.

Plane A:

tan(13°) = 10,000/h

h•tan(13°) = 10,000

h = 10,000/tan(13°) = 43,315 feet

Plane B:

tan(9°) = 10,000/h

h•tan(9°) = 10,000

h = 10,000/tan(9°) = 63,138 feet

Correct answer:

Plane B because it was 63,138 feet away.

User Anthony Pham
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8.2k points