To approach the runway, a pilot of a small plane must begin a 20° descent starting from a height of 3,760 feet above the ground. To the nearest tenth of a mile, how many miles from the runway is the airplane at the start of this approach? The figure is not drawn to scale. (1 point) Responses 10,993.5 mi 10,993.5 mi 2.1 mi 2.1 mi 1.8 mi 1.8 mi 0.8 mi 0.8 mi Skip to navigation

To solve this problem, we can use trigonometry.

First, we need to calculate the altitude the plane will descend to when it reaches the runway. Since the plane is beginning a 20° descent from a height of 3,760 feet, the vertical displacement can be calculated as:

Vertical displacement = 3,760 ft * sin(20°)
Vertical displacement = 3,760 ft * 0.3420
Vertical displacement ≈ 1,286.32 ft

Next, we need to calculate the horizontal distance the plane will travel during this descent. This can be calculated as:

Horizontal distance = 3,760 ft * cos(20°)
Horizontal distance = 3,760 ft * 0.9397
Horizontal distance ≈ 3,536.55 ft

To find the distance from the runway to the plane at the start of the approach, we can use the Pythagorean theorem:

Distance = √(Horizontal distance^2 + Vertical displacement^2)
Distance = √(3,536.55^2 + 1,286.32^2)
Distance ≈ √(12,511,406.25 + 1,652,736.94)
Distance ≈ √14,164,143.19
Distance ≈ 3,765.28 ft

Converting this distance to miles, we have:

3,765.28 ft ÷ 5,280 ft/mile ≈ 0.71 miles

Therefore, the airplane is approximately 0.7 miles from the runway at the start of the approach. The closest answer choice is 0.8 mi.