Question

Can someone please help

Answer 1

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 = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

sinθ = $\frac{AC}{AB}$

       = $\frac{1500}{5000}$

$\theta=\text{sin}^{-1}(\frac{3}{10})$

$\theta=17.46$ degrees