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An airplane is flying at an altitude of 6000 m over the ocean directly toward acoastline. At a certain time, the angle of depression to the coastline from the airplaneis 14°. How much farther (to the nearest meter) does the airplane have to fly before itis directly above the coastline?O 5822 mO 1496 mO 1451 mO 24065 m

An airplane is flying at an altitude of 6000 m over the ocean directly toward acoastline At a certain time the angle of depression to the coastline from the air class=

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Explanation

Using the data of the statement, we make the following diagram:

From the diagram, we see that:

• ΔPSC forms a right triangle,

,

• m∠C = θ = 14° is the angle of depression,

,

• oc = PS = 6000 m is the opposite side to angle θ,

,

• ac = SC = x is the adjacent side to angle θ.

From trigonometry, we have the relation:

[tex]\tan\theta=\frac{oc}{ac}\Rightarrow ac=\frac{oc}{\tan\theta}.[/tex]

Replacing the data from above, we get:

[tex]x=\frac{6000\text{ }m}{\tan(14\degree)}\cong24065\text{ }m.[/tex]Answer

24065 m

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