Data:
h (Tower height) = 120 ft
d (Distance from the base of the tower to the airplane) = ?
[tex] \alpha [/tex] (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:
[tex]tg \alpha = \frac{opposite\:leg}{adjacent\:leg} [/tex]
Solving:
[tex]tg \alpha = \frac{opposite\:leg}{adjacent\:leg} [/tex]
[tex]tg 19^0 = \frac{h}{d} [/tex]
[tex]tg 19^0 = \frac{120}{d} [/tex]
[tex]0.34 = \frac{120}{d} [/tex]
[tex]0.34*d = 120[/tex]
[tex]0.34d = 120[/tex]
[tex]d = \frac{120}{0.34} [/tex]
[tex]\boxed{\boed{d \approx 352.9\:ft}}\end{array}}\qquad\quad\checkmark[/tex]
Answer:
Distance from the base of the tower to the airplane is aproximately 352.9 ft
