Question

Two ships are anchored on opposite sides of a lighthouse. Their angles of depression as observed from the top of the lighthouse are found to be $30°$ and $45°$. The line joining the ships passes through the foot of the lighthouse. If the height of the lighthouse is $100\mathrm{m}$, find the distance between the ships.

Answer 1

Step 1: Drawing the diagram of the situation

Let $AB$ be a lighthouse with height $100\mathrm{m}.$

Let $C$ and $D$ be the position of two ships.

And their angles of depression as observed from the top of the lighthouse are found to be $30°$ and $45°$.

So, $\angle FAC=\angle ACB=30°$ and $\angle EAD=\angle ADB=45°$ (Alternate angles)

Let $CB=x$and $BD=y$ meters such that the distance between two ships will be $x+y$ meters.

Step 2. Find the value of $x$ and $y$

In $\Delta ABC$,

$\tan 30°=\frac{AB}{BC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{x}\left(\because \tan 30°=\frac{1}{\sqrt{3}}\right)$

$\Rightarrow x=100\sqrt{3}\mathrm{m}$

And

In $\Delta \mathrm{ABD}$,
$$\begin{gathered} \tan 45°=\frac{AB}{BD} \\ \Rightarrow 1=\frac{{\displaystyle 1000}}{{\displaystyle y}}\left(\because \tan 45°=1\right) \\ \Rightarrow y=100\mathrm{m} \end{gathered}$$

Step 3: Finding the distance between the ships

Since the distance between two ships is $x+y$ meters.

$$\begin{gathered} x+y=100\sqrt{3}+100 \\ =100(\sqrt{3}+1) \end{gathered}$$

$=273.2051\mathrm{m}$

Hence, the distance between two ships is $273.20$ meters.