Question

A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point $b$ metres just above A is $\beta$. Prove that the height of the tower is $b\tan \alpha \cot \beta$.

Answer 1

Here, $CD$ is the height of the tower.

In $△ABC,\tan \beta =\frac{AB}{AC}$

$\Rightarrow$ $\tan \beta =\frac{b}{AC}$

$\therefore$ $AC=\frac{b}{\tan \beta }$

$\therefore$ $AC=b\cot \beta$ ------ ( 1 )

In $△ACD,\tan \alpha =\frac{CD}{AC}$

$\Rightarrow$ $CD=\tan \alpha \times AC$

$\Rightarrow$ $CD=\tan \alpha \times b\cot \beta$ [ From ( 1 ) ]

$\therefore$ $CD=b\tan \alpha \cot \beta$ [ Hence proved ]