Question

From a lighthouse, the angles of depression of two ships on opposite sides of the lighthouse were observed to be $30°$ and $45°$ If the lighthouse is $90m$ high and the line joining the two ships passes through the foot of the lighthouse, find the distance between the two ships.

Answer 1

Angle of depression : The angle of depression is the angle between the horizontal line and the line of observation.

Step 1: Drawing the diagram of the situation

Let $AB$ be the lighthouse of height $90m.$

And let $C$ and $D$ be the positions of two ships such that angles of depression $A\text{'}AD$ and $A\text{'}\text{'}AC$ are $30°$ and $45°$ respectively.

So, $\angle A\text{'}AD=30°$ and $\angle A\text{'}\text{'}AC=45°.$

From the figure, we have

$\angle BAC=45°$ and $\angle BAD=60°.$

So that, $\angle ADB=30°$ and $\angle ACB=45°.$

Step 2: Finding the distance between the two ships

In right-angled $△ABC$

$$\begin{gathered} \tan \left(C\right)=\frac{\text{side opposite to}\angle C}{\text{side adjacent to}\angle C} \\ \Rightarrow \tan \left(45°\right)=\frac{AB}{BC} \\ \Rightarrow 1=\frac{90}{BC}(\because \tan \left(45°\right)=1) \\ \Rightarrow BC=90m \end{gathered}$$

Again, in right-angled $△ABD$

$$\begin{gathered} \tan \left(D\right)=\frac{\text{side opposite to}\angle D}{\text{side adjacent to}\angle D} \\ \Rightarrow \tan \left(30°\right)=\frac{AB}{BD} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{90}{BD}\left(\because \tan \left(30°\right)=\frac{1}{\sqrt{3}}\right) \\ \Rightarrow BD=90\sqrt{3}m \end{gathered}$$

The distance between the two ships $DC=BD+BC$

$$\begin{gathered} =90\sqrt{3}+90 \\ =155.88+90 \\ =245.88m \end{gathered}$$

Therefore, the distance between the two ships is $245.88m$.