Answer:
Angle of elevation of the airplane = 17.46 degrees
Step-by-step explanation:
From the picture attached,
An airplane is flying at an altitude of 1500 ft at point A.
Runway starts from point B from which distance of the airplane is 5000 ft.
Now we apply sine rule in the given triangle ABC to measure the angle θ.
sinθ = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]
sinθ = [tex]\frac{AC}{AB}[/tex]
= [tex]\frac{1500}{5000}[/tex]
[tex]\theta=\text{sin}^{-1}(\frac{3}{10})[/tex]
[tex]\theta=17.46[/tex] degrees
