Question

A person on a runway sees a plane approaching. The angle of elevation from the runway to the plane is 11.1° . The altitude of the plane is 4000 fr. What is the horizontal distance from the plane to the person on the runway.

Answer 1

Answer:

The horizontal distance from the plane to the person on the runway is 20408.16 ft.

Step-by-step explanation:

Consider the figure below,

Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner.  The angle of elevation from the runway to the plane is 11.1°

BC is the horizontal distance from the plane to the person on the runway.

We have to find distance BC,

Using trigonometric ratio,

$\tan\theta=\frac{Perpendicular}{base}$

Here, $\theta=11.1^{\circ}$ ,Perpendicular AB = 4000

$\tan\theta=\frac{AB}{BC}$

$\tan 11.1^{\circ} =\frac{4000}{BC}$

Solving for BC, we get,

$BC=\frac{4000}{\tan 11.1^{\circ} }$

$BC=\frac{4000}{0.196}$ (approx)

$BC=20408.16$(approx)  

Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft