Answer: The ship is 100√3 m (about 173.2 m) from the light-house.
Let AB = 100 m be the light-house and C the ship. The angle of depression \(30^\circ\) equals the angle of elevation of the top from the ship (alternate angles), so \(\angle ACB=30^\circ\).
\[\tan30^\circ=\frac{AB}{BC}\Rightarrow\frac1{\sqrt3}=\frac{100}{BC}\]
\[BC=100\sqrt3\approx173.2\ \text{m}\]