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

8 0

3 years ago

What is an equation of the vertical line that passes through (-2, -9)

ser-zykov [4K]

Answer: x = -2

Step-by-step explanation: vertical lines are the x axis

3 0

3 years ago

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