Question

The pilot of an aircraft flying horizontally at a speed 1200km/hr.Observes that the angle of depression of a point on the ground changes from ${30}^{\circ }$ to ${45}^{\circ }$ in 15 sec.Find the height at which the aircraft is flying.

Answer 1

REF.Image.

Solution:-

Speed of an aircraft $=1200km/hr$

After 15 secs, angle of depression changes from ${30}^{\circ }$ to ${45}^{\circ }$

Distance covered in 15 seconds $=1200\times 15\times 3600=5km=5000m$

Let height from ground be $h$

Horizontal distance $=x$

In $△BDP,\angle P={45}^{\circ },\therefore tan{45}^{\circ }=\frac{BD}{DP}=\frac{h}{DP}$

$\therefore DP=h$

In $△ACP,\angle APC={30}^{\circ }$

$\therefore tan{30}^{\circ }=\frac{AC}{CP}$

$\therefore \frac{1}{\sqrt{3}}=\frac{h}{5000+h}$

$\Rightarrow 5000+h=h\sqrt{3}$

$\Rightarrow 5000=h(\sqrt{3}-1)$

$\Rightarrow h=\frac{5000}{0.732}$

$\Rightarrow h=6830m=6.830km$ (height of an aircrft from ground)