Answer to Question #340571 in Trigonometry for bmully

Question

an airplane is flying at an altitude of 6000m over the ocean directly toward a coastline. at a certain time the angle of depression to the coastline from the airplane is 14 degrees. how much farther does the airplane have to fly before it is directly over the coastline

Answer 1

In a triangle of the plane, the coast line and the point where the plane is over the coast line, the angle at

the plane is 14° and the side across from this angle (distance from coastline straight up to point where

plane is above it) is 6000 𝑚, so if the diagonal line from the plane to the coastline is 𝑥, we know that

𝑥 sin 14° = 6000 𝑚, or 𝑥 = 6000/sin 14°

We are looking for the length of the third side , let's call that 𝑦 and we

know that 𝑥 cos 14° = 𝑦 so

𝑦 = ( 6000/sin 14°) cos 14° = 6000 cot 14° = 6000 ∙ 4.01 = 24060 𝑚 = 24 km