Question

An aeroplane when flying at a height of 4000m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of two planes from the same point on the ground are 60° and 45°. Find the vertical distance between the aeroplane at that instant..

Answer 1

Final answer is 1690.59m


Hope it helps......

Answer 2

Answer:

1690.599 m

Step-by-step explanation:

Refer the attached figure

Height of first Airplane = AB = 4000 m

Height of another plane = BC

The angles of elevation of two planes from the same point on the ground are 60° and 45°. i.e.∠ADB = 60° and ∠CDB = 45°

Let AC be h

CB = AB - AC = 4000-h

In ΔABD

$tan \theta = \frac{Perpendicular}{Base}$

$tan 60^{\circ} = \frac{AB}{BD}$

$\sqrt{3}= \frac{4000}{BD}$

$BD= \frac{4000}{\sqrt{3}}$

$BD= 2309.401$

In ΔCBD

$tan \theta = \frac{Perpendicular}{Base}$

$tan 45^{\circ} = \frac{CB}{BD}$

$1= \frac{4000-h}{2309.401}$

$2309.401=4000-h$

$h=4000-2309.401$

$h=1690.599$

Hence the vertical distance between the aeroplane at that instant is 1690.599 m