Question

The angle of elevation of an aeroplane from a point on the ground is ${45}^{\circ }$. After a flight of 15 seconds, the elevation changes to ${30}^{\circ }$. If the aeroplane is flying at a height of 3000 meters, find the speed of the aeroplane.

Answer 1

Sol:



Let the initial position of the plane be at A and it moves to the next position at D in 15 sec.
Let BCE be the horizontal line on the ground. B is the point of observation. Two angles of elevation of the plane from the point B are ∠ABC = 45° and ∠DBC = 30°.
ΔABC forms a right angled triangle with AB as the hypotenuse, BC as the base and AC as the height.
AC = height of the plane = 3000 m

$tan45=\frac{AC}{BC}$

$1=\frac{3000}{BC}$

$BC=3000$

Similarly, ΔDBE forms a right angled triangle with BD as the hypotenuse, BE as the base and DE as the height.
DE = height of the plane = 3000 m

$tan30=\frac{DE}{BE}$

$\frac{1}{\sqrt{3}}=\frac{3000}{BE}$

$BE=5196.15m$

Distance travelled by the plane = BE – BC = 5196.15 – 3000 = 2196.15 m

Time taken to travel = 15 sec

Speed of the plane = $\frac{2196.15}{15}$ = 146.41 m/sec.