Question

From a helicopter vertically above a straight horizontal plane. If the angles of depression of two consecutive kilometer stones on the same side of the helicopter are found to be θ and φ, where θ > φ, then the height of the helicopter from the ground is (in km)

• A. $\sqrt{\tan \theta \tan \varphi }$
• B. $\frac{\tan \theta \tan \varphi }{\tan \theta -\tan \varphi }$
• C. $\frac{\tan \theta -\tan \varphi }{2(\tan \theta +\tan \varphi )}$
• D. $\frac{\tan \theta \tan \varphi }{(\tan \theta +\tan \varphi )}$

Answer 1

The correct option is B $\frac{\tan \theta \tan \varphi }{\tan \theta -\tan \varphi }$

Let AB be the height of the helicopter from the ground.

$In\Delta ABC,$
$\tan \theta =\frac{AB}{BC}$
$\Rightarrow BC=\frac{AB}{tan\theta }$ ....(i)
$In\Delta ABD,$
$\tan \varphi =\frac{AB}{DB}$
$\Rightarrow \varphi =\frac{AB}{DC+BC}$ $(\because DB=DC+BCandDC=1km)$
$\Rightarrow \tan \varphi =\frac{AB}{1+BC}$ $(\because DC=1km)$
$\Rightarrow 1+BC=\frac{AB}{\tan \varphi }$
$\Rightarrow 1+\frac{AB}{\tan \theta }=\frac{AB}{\tan \varphi }$ [From (i)]
$\Rightarrow 1=\frac{AB}{\tan \varphi }-\frac{AB}{\tan \theta }$
$\Rightarrow 1=\frac{AB(\tan \theta -\tan \varphi )}{\tan \theta \tan \varphi }$
$\Rightarrow AB=\frac{\tan \theta \tan \varphi }{\tan \theta -\tan \varphi }$
∴ Required height of the helicopter is AB = $\frac{\tan \theta \tan \varphi }{\tan \theta -\tan \varphi }$
Hence, the correct answer is option (b).