USE ESTIMATION The angle of elevation to an airplane viewed from the air traffic control tower at an airport is . The tower is 200 feet tall, and the pillot roports that the altitude of the aplane is 5127 feet.
2. Explain how you can use angles of elevation and depression to estimate the distance from the air traffic control tower to the airplane.
and the height of the air traffic control tower are known, I can find the
Then, I can use this Select Choice and the angle of elevation to etimate the distance.

Part B
Incorrect
2 tries left. Please try again.
b. About how far away is the airplane from the air traffic control uower be the neved low
f

Ask by Parry Guerrero. in the United States

Mar 28,2025

Question

USE ESTIMATION The angle of elevation to an airplane viewed from the air traffic control tower at an airport is . The tower is 200 feet tall, and the pillot roports that the altitude of the aplane is 5127 feet.
2. Explain how you can use angles of elevation and depression to estimate the distance from the air traffic control tower to the airplane.
and the height of the air traffic control tower are known, I can find the
Then, I can use this Select Choice and the angle of elevation to etimate the distance.

Part B
Incorrect
2 tries left. Please try again.
b. About how far away is the airplane from the air traffic control uower be the neved low
f

Ask by Parry Guerrero. in the United States

Mar 28,2025

UpStudy AI · Accepted Answer

1. Compute the vertical height difference between the airplane and the tower:
   $$
   \text{Vertical height} = 5127 - 200 = 4927 \text{ feet.}
   $$

2. In the right triangle formed by the tower, the airplane, and the horizontal distance between them, the $$7^\circ$$ angle of elevation has a tangent given by:
   $$
   \tan 7^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{4927}{d},
   $$
   where $$ d $$ is the horizontal distance from the tower to the airplane.

3. Solve for $$ d $$ by rearranging the formula:
   $$
   d = \frac{4927}{\tan 7^\circ}.
   $$

4. Estimate $$\tan 7^\circ$$. Noting that $$\tan 7^\circ \approx 0.1228$$, we have:
   $$
   d \approx \frac{4927}{0.1228} \approx 40\,000 \text{ feet.}
   $$

5. To explain the use of angles of elevation and depression: Since the tower’s height and the airplane’s altitude are known, you can subtract these to find the vertical distance between them. Then, using the angle of elevation (or the corresponding angle of depression from the airplane, which is equal by alternate interior angles), you can set up a tangent ratio. This tangent ratio relates the vertical distance (opposite side) with the horizontal distance (adjacent side) of the right triangle. Solving for the horizontal distance estimates how far the airplane is from the tower.

Thus, the airplane is approximately $$40\,000$$ feet away from the air traffic control tower.
The airplane is about 40,000 feet away from the air traffic control tower.

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