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:

Solution(By Examveda Team)

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