A tower subtends an angle $\alpha$ on the same level as the foot of the tower and at a second point $h$ meters above the first, the depression of the foot of the tower is $\beta$. Show that the height of the tower is $h \tan \alpha \cot \beta$.
In the figure,
$AT = $ tower of height $H$
$B = $ a point on the same level as the foot of the tower at a distance $x$ from $A$
$C = $ a point $h$ meters above $B$
From the figure,
in $\triangle ATB$, $\;$ $\tan \alpha = \dfrac{AT}{AB} = \dfrac{H}{x}$
in $\triangle ACB$, $\;$ $\tan \beta = \dfrac{BC}{AB} = \dfrac{h}{x}$
$\therefore \;$ $\dfrac{\tan \alpha}{\tan \beta} = \dfrac{H / x}{h / x} = \dfrac{H}{h}$
i.e. $\;$ $H = \dfrac{\tan \alpha}{\tan \beta} \times h$
i.e. $\;$ $H = h \tan \alpha \cot \beta$