The angle of depression of two ships from an aeroplane flying at the height of
7500 𝑚 are 30$°$ and 45$°$ . If both the ships are in the same line that one ship is exactly behind the other, find the side such distance between the ships.

We are given the height at which the aeroplane is flying and the angle of depressions it makes with the two ships.

The height at which the aeroplane is flying, ℎ = 7500 𝑚

The angle of depression of the first ship, θ = 30$°$

And the angle of depression of the second ship, θ' = 45$°$

The figure depicting this condition is shown in the given figure,

Here, 𝑥 is the distance between the two ships. In ∆𝐴𝐵𝐶, ∠𝐵 is the right angle.
So, 𝑡𝑎𝑛 θ' =   𝐴𝐵/BC
⇒ 𝑡𝑎𝑛 45$°$ = 7500/𝑦
⇒ 1 = 7500 
⇒ 𝑦 = 7500 𝑚
Now, in ∆𝐴𝐵𝐷, ∠𝐵 is the right angle. So, 𝑡𝑎𝑛 θ =    𝐴𝐵/BD
⇒ 𝑡𝑎𝑛 30$°$ =7500/𝑥+𝑦

⇒    1/$\sqrt{3}$ =    7500/𝑥+7500
⇒ 𝑥 + 7500 = 7500 × $\sqrt{3}$
⇒ 𝑥 + 7500 = 7500 × 1. 73
⇒ 𝑥 + 7500 = 12975
⇒ 𝑥 = 12975 − 7500
⇒ 𝑥 = 5475 𝑚
Hence, the distance between the two ships is 5475 𝑚