Question

An airplane is flying at an altitude of 6000 m over the ocean directly towards a coastline. At a certain time, the angle of depression to the coastline from the airplane is 14 degrees. How much farther ( to the nearest kilometre ) does the airplane have to fly before it is directly above the coastline?

Answer 1

Answer on Question #44186 – Math - Trigonometry

An airplane is flying at an altitude of $6000\,\mathrm{m}$ over the ocean directly towards a coastline. At a certain time, the angle of depression to the coastline from the airplane is 14 degrees. How much farther ( to the nearest kilometer ) does the airplane have to fly before it is directly above the coastline?

Solution

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 $14{}^{\circ}$ and the side across from this angle (distance from coastline straight up to point where plane is above it) is $6000\,\mathrm{m}$, so if the diagonal line from the plane to the coastline is $x$, we know that $x\sin 14{}^{\circ} = 6000\,\mathrm{m}$, or $x = \frac{6000}{\sin 14{}^{\circ}}$. We are looking for the length of the third side, let's call that $y$ and we know that $x\cos 14{}^{\circ} = y$ so

$$y = \left(\frac{6000}{\sin 14{}^{\circ}}\right) \cos 14{}^{\circ} = 6000 \cot 14{}^{\circ} = 6000 \cdot 4.01 = 24060\,\mathrm{m} = 24\,\mathrm{km}.$$

Answer: 24 km.

www.AssignmentExpert.com