Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 173 m B. 200 m C. 273 m D. 300 m Answer: Option CShow Answer
Solution(By Apex Team)
Let, BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,
$\angle B A D=30^{\circ}, \angle B C D=45^{\circ}$
$\begin{array}{l}
\tan 30^{\circ}=\frac{B D}{B A} \\
\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{B A} \\
\Rightarrow B A=100 \sqrt{3} \\
\tan 45^{\circ}=\frac{B D}{B C} \\
\Rightarrow 1=\frac{100}{B C} \\
\Rightarrow B C=100
\end{array}$
Distance between the two ships;
$\begin{array}{l}
=A C \\
=B A+B C \\
=100 \sqrt{3}+100 \\
=100(\sqrt{3}+1) \\
=100(1.73+1) \\
=100 \times 2.73=273 m
\end{array}$
