A   is the aeroplane, D  and  C  are the ships sailing towords  A.

Ships are sailing towords the aeroplane in the same direction.

in the figure, height  Ab = 1500 m.

To find:  Distance between the ships, that is  CD.

Solution: 

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{right}-\mathrm{angled} ∆\mathrm{ABC}, \\ \tan 45° = \frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1 = \frac{1500}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC} = 1500 \mathrm{m}. \\ \mathrm{In} \mathrm{the} \mathrm{right}-\mathrm{angled} ∆\mathrm{ABD}, \\ \tan 30° = \frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD} = 1500 \sqrt{3} \\ \Rightarrow \mathrm{BD} = 1500 ( 1.732 ) \\ = 2598 \mathrm{m} \\ \therefore \mathrm{Distance} \mathrm{between} \mathrm{the} \mathrm{ships} = \mathrm{CD} = \mathrm{BD} - \mathrm{BC} \\ = 2598 - 1500 \\ = 1098 \mathrm{m}. \end{gathered}$$