From the top of a tower $ h $ metre high, the angles of depression of two objects, which are in the line with the foot of the tower are $ \alpha $ and $\beta(\beta>\alpha) $. Find the distance between the two objects.

Given:

From the top of a tower \( h \) metre high, the angles of depression of two objects, which are in the line with the foot of the tower are \( \alpha \) and \(\beta(\beta>\alpha) \).

To do:

We have to find the distance between the two objects.

Solution:

Let the distance between the two objects be $x\ m$, $DA= y\ m$ and the height of the tower be $AB = h\ m$.

From the figure,

$\angle XBC=\angle BCA = \alpha$                        (Alternate angles are equal)
$\angle XBD = \angle BDA = \beta$                            (Alternate angles are equal)
 In $\Delta \mathrm{ABD}$,

$\tan \beta=\frac{\mathrm{AB}}{\mathrm{AD}}$

$=\frac{h}{y}$

$\Rightarrow y=\frac{h}{\tan \beta}$.............(i)

In $\Delta \mathrm{BCA}$,

$\tan \alpha=\frac{\mathrm{AB}}{\mathrm{AC}}$

$\Rightarrow \tan \alpha=\frac{\mathrm{AB}}{\mathrm{CD}+\mathrm{AD}}$

$\Rightarrow \tan \alpha=\frac{h}{x+y}$

$\Rightarrow x+y=\frac{h}{\tan \alpha}$

$\Rightarrow y=\frac{h}{\tan \alpha}-x$............(ii)

From equations (i) and (ii),

$\frac{h}{\tan \beta}=\frac{h}{\tan \alpha}-x$

$x=\frac{h}{\tan \alpha}-\frac{h}{\tan \beta}$

$x=h(\frac{1}{\tan \alpha}-\frac{1}{\tan \beta})$

$x=h(\cot \alpha-\cot \beta)$

Therefore, the distance between the two objects is $h(\cot \alpha-\cot \beta)$.

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Updated on: 10-Oct-2022

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