Question

from the top of a 120-foot high tower an air traffic controller observes an airplane on the runway at an angle of depression of 19 degrees how far from the base of the tower is the airplane

Answer 1

Data:
h (Tower height) = 120 ft
d (Distance from the base of the tower to the airplane) = ?
$\alpha$ (Angle formed from observation of the tower to the airplane)=19º

Data: tg 19º ≈ 0.34

Note: The angle formed is tangent to the height of the tower and the distance from the base of the tower to the airplane.

Formula:
$tg \alpha = \frac{opposite\:leg}{adjacent\:leg}$

Solving:
$tg \alpha = \frac{opposite\:leg}{adjacent\:leg}$
$tg 19^0 = \frac{h}{d}$
$tg 19^0 = \frac{120}{d}$
$0.34 = \frac{120}{d}$
$0.34*d = 120$
$0.34d = 120$
$d = \frac{120}{0.34}$
$\boxed{\boed{d \approx 352.9\:ft}}\end{array}}\qquad\quad\checkmark$

Answer:
Distance from the base of the tower to the airplane is aproximately  352.9 ft