Question

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

Answer 1

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/√3 = 7500/x + 7500

⇒ x + 7500 = 7500 × √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.

Answer 2

Hence, the distance between two ships

= 5475 m