Answer:
The distance between the runway and the airplane where it start this approach is 2.16 miles.
Step-by-step explanation:
A pilot of a small plane must begin a 10 descent starting from a height of 1983 feet above the ground that is AB is the height of plane above the ground, AB= 1983 feet. and A is the point from where the pilot starts descent.
thus, ∠ACB = ∠DAC = 10°
We have to find the distance between the runway and the airplane where it start this approach that is we have to find length AC( in miles) . Let AC = x
Applying trigonometric ratio,
\(\sin C=\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\)
Substitute the values,
\(\sin 10^{\circ}=\dfrac{1983}{x}\)
\(x=\dfrac{1983}{\sin 10^{\circ}}\)
\(x=11422.811\)(approx)
Thus, Distance between the runway and the airplane is 11422.81 feet.
Since , we have to find distance in miles
We know 1 mile = 5280 feet
1 feet = \(\frac{1}{5280}\) miles.
11422.81 feet = \(\frac{1}{5280}\times 11422.81\) miles.
= 2.16 miles (approx)
Thus, the distance between the runway and the airplane where it start this approach is 2.16 miles.
