Question

An aeroplane at an altitude 1500m, find that the two ships are sailing towards it in the same direction. The angle of depression as observed from the airplane is \[{{45}^{\circ }}\] and \[{{30}^{\circ }}\] respectively. Find the distance (in m) between two ships.

Answer 1

Hint: Now first note that the ships and the aeroplane will form two right-angle triangles. Now we know the angle of depression of ships, one angle of the triangle is 90 degrees. Now we will use the tan ratio defined as $\tan \theta =\dfrac{opposite}{adjecent}$ on each angle of depression. Using this we will get an equation to find the distance between ships.

Complete step-by-step solution:
Now first let us understand the Question. It is given that an aeroplane at an altitude 1500m hence the vertical distance is 1500m let the Aeroplane be represented by A. Now the two ships are sailing towards it in the same direction. The angle of depression as observed from the airplane is \[{{45}^{\circ }}\] and \[{{30}^{\circ }}\] respectively let this be C and D respectively. So let us draw the figure for this

Now in the figure triangle, ABC and triangle ACB is a right angle triangle.
Now first Consider triangle ABC
In this triangle we have $\angle ABC={{90}^{\circ }},\angle BAC={{45}^{\circ }}$ and we have AB = 1500m
Now consider tan of angle BAC.
We know that tan ratio is defined as $\tan \theta =\dfrac{opposite}{adjecent}$
Hence we get
$\begin{align}
  & \tan \angle BAC=\dfrac{BC}{AB} \\
 & \tan {{45}^{\circ }}=\dfrac{BC}{1500} \\
\end{align}$
Now we know that $\tan {{45}^{\circ }}$ is 1
Hence we get
$\begin{align}
  & 1=\dfrac{BC}{1500} \\
 & \Rightarrow BC=1500m............................\left( 1 \right) \\
\end{align}$
Now consider the triangle ABD.
Again in this triangle we have $\angle ABD={{90}^{\circ }},\angle DAB={{30}^{\circ }}$and AB = 1500m
Now consider tan of $\angle DAB$
We know that tan angle is defined as $\tan \theta =\dfrac{opposite}{adjecent}$
\[\tan \angle DAB=\tan {{30}^{\circ }}=\dfrac{BD}{AB}\]
Now we know that $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ and AB = 1500m.
$\begin{align}
  & \dfrac{1}{\sqrt{3}}=\dfrac{BD}{1500} \\
 & \Rightarrow BD=\dfrac{1500}{\sqrt{3}} \\
\end{align}$
Now rationalizing we get
$BD=\dfrac{1500\sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\dfrac{1500\sqrt{3}}{3}=500\sqrt{3}$
Hence we have $BD=500\sqrt{3}m...............................\left( 2 \right)$
Now we have from figure that the distance between Ship C and ship D is
BC – BD. Let this distance be d, then we have
$\begin{align}
  & d=BC-BD \\
 & \Rightarrow d=1500m-500\sqrt{3}m \\
\end{align}$
Now we know that value of $\sqrt{3}=1.73$ hence we get.
$\begin{align}
  & d=1500-500\left( 1.73 \right) \\
 & \Rightarrow d=1500-865 \\
 & \Rightarrow d=635m \\
\end{align}$
Hence the distance between the ships is 635m.

Note: Note that here we have taken tan ratio since we wanted to find the opposite side and we have the value of the adjacent side. In such problems always choose your trigonometric ratio accordingly. Also, note in the question angle of depression is given and not the angle of elevation. Hence we will solve accordingly.