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 200 m high, the distance between the two ships is:

Answer: Option C

Solution

Answer: Option C

Solution:

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 200 m, ∠BAD = 30°, ∠BCD = 45°
$$\begin{array}{rl}& \tan 30^{\circ }=\frac{BD}{BA}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{200}{BA}\\ & \Rightarrow BA=200\sqrt{3}\\ & \tan 45^{\circ }=\frac{BD}{BC}\\ & \Rightarrow 1=\frac{200}{BC}\\ & \Rightarrow BC=200\end{array}$$
Distance between the two ships
$$\begin{array}{rl}& =AC=BA+BC\\ & =200\sqrt{3}+200\\ & =200(\sqrt{3}+1)\\ & =200\left(1.73+1\right)\\ & =200\times 2.73\\ & =546m\end{array}$$