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. 85 (√3 – 1) m. B. 65 (√2 – 1) m. C. 75 (√3 – 1) m. D. 55 (√3 – 1) m. Answer: Option CShow Answer
Solution(By Apex Team)
Let the ships be A and B and OH be the light house
$\begin{array}{l}\therefore\tan45^{\circ}=\frac{\text{AB}}{\text{BC}}\\
\text{AB}=\text{BC}\\
\text{BC}=75\ \text{m}\\
\therefore\tan30^{\circ}=\frac{\text{AB}}{\text{BD}}\\
\frac{1}{\sqrt{3}}=\frac{\text{AB}}{\text{BD}}\\
\text{BD}=\text{AB}\sqrt{3}\\
=75\sqrt{3}m\end{array}$
Distance between ships is CD;
$\begin{array}{l}\therefore\text{CD}=\text{BD}-\text{BC}\\
=75\sqrt{3}-75\ \text{m}\\
=75(\sqrt{3}-1)\ \text{m}\\
\text{CD}=54.90\ \text{m}\end{array}$
EN 
HI 