Question

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.

Answer 1

Let C and D be the two aeroplanes and A be the point of observation. Then,

∠CAB = 30°, ∠DAB = 60°, BC = 3125 m

Let DC = y m, AB = x m

In right ΔABC, ∠B = 90°

tan 30° = `("BC")/("AB")`

⇒ `1/sqrt(3) = 3125/("AB")`

⇒ AB = `3125sqrt(3)` m  ...(i)

In right ΔABD, ∠B = 90°

tan 60° = `("BD")/("AB")`

⇒ `sqrt(3) = ("y"+ 3125)/(3125 sqrt(3))`  ...[From equation (i)]

⇒ 3125 × 3 = y + 3125

⇒ y = 3125 (3 – 1)

⇒ y = 2 × 3125

⇒ y = 6250 m

Therefore, the distance between the two planes is 6250 m.