As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
A. 20√3m. B. 18√3m. C. 16√3m. D. 22√3m. Answer: Option AShow Answer
Solution(By Apex Team)
In Right triangle,
$\Delta C D O$
$\begin{array}{l}\tan30^{\circ}=\frac{CD}{OD}\\
\Rightarrow\frac{1}{\sqrt{3}}=\frac{CD}{OD}\\
\Rightarrow CD=\frac{OD}{\sqrt{3}}\ldots(i)\end{array}$
Also, In Right Angle;
$\triangle A B O,$
$\begin{array}{l}
\tan 60^{\circ}=\frac{A B}{O B} \\
\Rightarrow \sqrt{3}=\frac{A B}{(80-O D)} \\
\Rightarrow A B=\sqrt{3}(80-O D) \\
A B=C D \\
\Rightarrow \sqrt{3}(80-O D)=\frac{O D}{\sqrt{3}} \\
\Rightarrow 3(80-O D)=O D \\
\Rightarrow 240-3 O D=O D \\
\Rightarrow 4 O D=240 \\
\Rightarrow O D=60 \mathrm{~m}
\end{array}$
Putting the value of OD in equation (i)
$\begin{array}{l}CD=\frac{OD}{\sqrt{3}}\\
\Rightarrow CD=\frac{60}{\sqrt{3}}\\
\Rightarrow CD=20\sqrt{3}m\end{array}$