We can solve this type of question using trigonometric functions. Here, we use sine of the angle of take-off with the horizontal runway.
[tex]\begin{gathered} \sin 10^o=\frac{500}{c} \\ c=\frac{500}{\sin 10^o} \\ c=\frac{500}{0.1736} \\ c=2880.18 \end{gathered}[/tex]
Thus, the distance that the plane has traveled is equal to 2880 ft.
