Question

Can someone please help

question: An airplane is flying at an elevation of 1500 feet. What is the airplane's angle of elevation from the runway when it is 5000 feet from the runway?

Answer 1

Angle of elevation of the airplane = 17.46 degrees

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 =
`(AC)/(AB)`

=
`(1500)/(5000)`

`\theta=\text{sin}^(-1)((3)/(10))`

`\theta=17.46` degrees