Let, height of light house from sea level (AB)=100m
Let, two ships be at the positions be C and D
In △ABC,
tan 45° = \(\frac{AB}{BC}\)
⇒ 1 = \(\frac{100}{BC}\)
⇒ BC = 100m
In △ABD,
tan 30° = \(\frac{AB}{BD}\)
⇒ \(\frac1{\sqrt3}\) = \(\frac{AB}{BD}\)
BD = AB × 3
=100 × √3
=173.2 [√3=1.732]
∴ CD = BD − BC
= 173.2 - 100
= 73.2m
∴ The distance between two ships 73.2 m