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**