Question : The angle of elevation of the top of a tower, vertically erected in the middle of a paddy field, from two points on a horizontal line through the foot of the tower are given to be $\alpha$ and $\beta(\alpha>\beta)$. The height of the tower is $h$ units. A possible distance (in the same unit) between the points is:
Option 1: $\frac{h(\cot\beta–\cot\alpha)}{\cos(\alpha+\beta)}$
Option 2: $h(\cot\alpha–\cot\beta)$
Option 3: $\frac{h(\tan\beta–\tan\alpha)}{\tan\alpha\tan\beta}$
Option 4: $h(\cot\alpha+\cot\beta)$
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Correct Answer: $h(\cot\alpha+\cot\beta)$
Solution :
$AD=h$ unit
In $\Delta ADB$,
$\tan\alpha=\frac{AD}{BD}$
$⇒\tan\alpha=\frac{h}{BD}$
$⇒BD=h\cot\alpha$ --------------------------------------(1)
In $\Delta ADC$,
$\tan\beta=\frac{AD}{DC}$
$⇒\tan\beta=\frac{h}{DC}$
$⇒DC=h\cot\beta$ -------------------------------------(2)
Now $BC=BD+DC$
$⇒BC=h\cot\alpha+h\cot\beta$
$\therefore BC=h(\cot\alpha+\cot\beta)$
Hence, the correct answer is $h(\cot\alpha+\cot\beta)$.
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