Question

From an airplane flying, vertically above a horizontal road, the angles of depression of two consecutive stones on the same side of the airplane are observed to be 30° and 60° respectively. The height at which the airplane is flying in km is :

• (A) $\frac{4}{\sqrt{3}}$
• (B) $\frac{\sqrt{3}}{2}$
• (C) $\frac{2}{\sqrt{3}}$
• (D) $2$

Answer 1

The Correct Option is B

Given: The angle of depression of two consecutive stones when observed from an airplane are ${30}^{\circ }$ and ${60}^{\circ }$.We have to find the height at which the airplane is flying.
Consider,

Let, $AB=hkm$
$BC=xkm$ and $CD=1km$
Consider, right angled $\mathrm{△}ABC$,
$\tan {60}^{\circ }=\frac{AB}{BC}$$\Rightarrow \sqrt{3}=\frac{h}{x}$$\Rightarrow x=\frac{h}{\sqrt{3}}$.........(i)
In right angled $\mathrm{△}ABD,$ we have
$\tan {30}^{\circ }=\frac{AB}{BD}$$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+1}$$\Rightarrow x+1=\sqrt{3}h$.......(ii)
From (i) and (ii), we get
$\frac{h}{\sqrt{3}}+1=\sqrt{3}h$$\Rightarrow h+\sqrt{3}=3h$$\Rightarrow 2h=\sqrt{3}$$\Rightarrow h=\frac{\sqrt{3}}{2}$ km
Hence, the correct option is (B).