An airplane is flying an altitude of 6000m over ocean. The angle of depression is 14 degrees to coastline. How much farther (in kilometers) does the plane have to fly before it is directly above the coastline?

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See attached image for the figure. The second link also provides a figure that illustrates an angle of depression.

Assuming that everything is just flat (no curvature on earth), the angle of depression from the plane is equal to the angle of elevation from the shore. This is because the path of the plane (broken line) is parallel to the horizontal ('ground') - and therefore, with the line of sight as the transversal, we have congruent alternate interior angles.

Then, we know that:

`tan theta = (opp)/(adj)`

We can calculate for the horizontal distance needed to be travelled by the plane.

`tan 14^o = (6000)/(adj)`

which means:

`adj = (6000)/(tan 14^o) = (6000)/(0.249) = 24064`

Hence, the plane needs to travel 24 064 meters.

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