Question

An Aeroplane flying horizontal and $1.2$ km above the ground is observed at an $\angle {60}^{\circ }$. After $15$sec the angle of elevation is observed to be ${30}^{\circ }$. Find the speed of aeroplane in km/hr. $(\sqrt{3}=1.73$)

Answer 1

Let say A is the initial position of the Aeroplan and B is the position of the Aeroplan after 15 sec. O is the observer position.

AB is the distance traveled by plane in 15 sec.

In $△OAC,$

$AC=1200m$

$tan{60}^o=\frac{AC}{OC}$

$\Rightarrow OC=\frac{1200}{\sqrt{3}}=693.64m$

In$△BOD,$

$tan{30}^o=\frac{BD}{OD}$

$\Rightarrow OD=\frac{1200}{\frac{1}{\sqrt{3}}}=2076m$

$AB=CD=OD-OC=2076-693.62=1382.36m$

$\therefore$ The speed of the aeroplane $=\frac{1382.36}{15}=92.16m/s$ $\Rightarrow \frac{92.16}{1000}\times 3600=331.76Km/hr$