The distances from the point the plane leaves the ground are given by the
trigonometric relationships of right triangle and Pythagoras theorem.
- A) The minimum distance to the base of the tower is approximately 874.57 ft.
- B) The minimum distance to the top of the tower is approximately 882.76 ft.
Reasons:
A) The angle with which the airplane climbs, θ = 11°
Height of the tower which the airplane flies, T = 120 foot
The clearance between the tower and the airplane, C = 50 feet
Required:
The minimum distance between the point where the plane leaves the ground and the base of the tower, [tex]\displaystyle d_{min}}[/tex]
Solution:
Height at which the plane flies over the tower, h = T + C
Therefore, h = 120 ft. + 50 ft. = 170 ft.
At the point the plane leaves the ground, we have;
- [tex]\displaystyle tan(\theta) = \mathbf{\frac{h}{d_{min}}}[/tex]
Which gives;
[tex]\displaystyle tan(11^{\circ}) = \frac{170 \, ft.}{d_{min}}[/tex]
[tex]\displaystyle d_{min} = \mathbf{\frac{170 \, ft.}{tan(11^{\circ})}} \approx 874.57 ft.[/tex]
- The minimum distance between the point where the plane leaves the ground and the base of the tower, [tex]\displaystyle d_{min}}[/tex] ≈ 874.57 ft.
B) The minimum distance between the point where the plane leaves the ground and the tower, R, is given by Pythagoras's theorem as follows;
R² = [tex]\displaystyle \mathbf{d_{min}}}[/tex]² + T²
Which gives;
R = √(*874.57 ft.)² + (120 ft.²)) ≈ 882.76 ft.
- The distance from the point where the airplane leaves the ground to the tower, R ≈ 882.76 ft.
Learn more about Pythagoras theorem here:
https://brainly.com/question/11256912
