Answer: The airport is 41043.6 feet far away from the plane.
Step-by-step explanation:
Since we have given that
A plane is flying at an altitude of 12,000 feet
As shown in the figure :
AB = 12000 feet
and the angle of elevation from the airport to the plane = 17 °
Consider, ΔABC ,
[tex]\sin 17^\circ=\frac{AB}{AC}\\\\\sin 17^\circ=\frac{12000}{AC}\\\\AC=\frac{12000}{\sin 17^\circ}\\\\BC=41043.64\ feet[/tex]
Hence, the airport is 41043.6 feet far away from the plane.
