Question

For a small plane, v , the angle of depression of a sailboat is 21 degrees. The angle of depression of a ferry on the other side of the plane is 52 degrees. The plane is flying at an altitude of 1650m how far apart are the boats, to the nearest meter?

Answer 1

Answer:

5,588 m

Explanation:

In the diagram:

$\begin{gathered} \angle\text{UVX}=90\degree-21\degree=69\degree \\ \angle\text{XVW}=90\degree-52\degree=38\degree \end{gathered}$

The distance between the two boats is UW and:

$UW=UX+XW$

In right triangle UXV:

$\begin{gathered} \tan V=\frac{UX}{VX} \\ \implies\tan 69\degree=\frac{UX}{1650} \\ \implies UX=1650\times\tan 69\degree \end{gathered}$

Similarly, in the right triangle WXV:

$\begin{gathered} \tan V=\frac{XW}{VX} \\ \implies\tan 38\degree=\frac{XW}{1650} \\ \implies XW=1650\times\tan 38\degree \end{gathered}$

Therefore:

$\begin{gathered} UW=UX+XW \\ =(1650\times\tan 69\degree)+(1650\times\tan 38\degree) \\ =5587.52m \\ \approx5,588m \end{gathered}$

The boats are 5,588 meters apart (correct to the nearest meter).