A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h meter. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and at the top of the flagstaff is β. Prove that the height of the tower is $\dfrac{\text{h tan α}}{\text{tan β} - \text{tan α}}$.

Let AB be the tower of height x meters, surmounted by a vertical flagstaff AD of height h meters (given). Let C be a point on the plane such that ∠ACB = α, ∠DCB = β and AD = h.

In ∆ABC,

$\Rightarrow \text{tan α} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan α}= \dfrac{AB}{BC} \\[1em] \Rightarrow BC = \dfrac{AB}{\text{tan α}} \\[1em] \Rightarrow BC = \dfrac{x}{\text{tan α}} ……….(1)$

In ∆DBC,

$\Rightarrow \text{tan β} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan β} = \dfrac{BD}{BC} \\[1em]$

Substituting value of BC from (1) in above equation :

$\Rightarrow \text{tan β} = \dfrac{BD}{\dfrac{x}{\text{tan α}}} \\[1em] \Rightarrow \text{tan β} = \dfrac{BA + AD}{\dfrac{x}{\text{tan α}}} \\[1em] \Rightarrow \text{tan β} \times \dfrac{x}{\text{tan α}} = x + h \\[1em] \Rightarrow x \text{tan β} = \text{tan α}(x + h) \\[1em] \Rightarrow x \text{tan β} = x \text{tan α} + h \text{tan α} \\[1em] \Rightarrow x \text{tan β} - x \text{tan α} = h \text{tan α} \\[1em] \Rightarrow x(\text{tan β} - \text{tan α}) = h \text{tan α} \\[1em] \Rightarrow x = \dfrac{\text{h tan α}}{\text{tan β} - \text{tan α}}.$

Hence, proved that the height of the tower = $\dfrac{\text{h tan α}}{\text{tan β} - \text{tan α}}.$