Question

The angle of elevation of an aeroplane from a point on the ground is 45 deg. after a flight of 15 seconds, the elevation changes to 30 deg, if the aeroplane is flying at a height of 3000 m find its speed

Answer 1

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Answer 2

Given :

The angle of elevation of an aeroplane from a point on the ground is 45 deg. after a flight of 15 seconds, the elevation changes to 30 deg,

To Find:

if the aeroplane is flying at a height of 3000 m find its speed

Solution:

Refer the attached figure

Height of Aeroplane = AB = 3000 m

The angle of elevation of an aeroplane from a point on the ground is 45 deg. i.e.∠ACB = 45°

So, $Tan \theta = \frac{Perpendicular}{Base}\\tan 45*{\circ}=\frac{AB}{BC}\\1=\frac{3000}{BC}\\BC=3000 m$

Let CD be x

So, BD = 3000+x

After a flight of 15 seconds, the elevation changes to 30 deg.

∠ADB=30°

$Tan \theta = \frac{Perpendicular}{Base}\\Tan 30^{\circ}=\frac{AB}{BD}\\\frac{1}{\sqrt{3}}=\frac{3000}{3000+x}\\3000+x=3000\sqrt{3}\\x=3000\sqrt{3}-3000\\Speed = \frac{Distance}{Time}\\Speed = \frac{3000\sqrt{3}-3000}{15}\\Speed=146.41$

Hence its speed is 146.41 m/s