A tower subtends an angle α on the same level as the foot of the tower and at a second point h metres above the first, the depression of the foot of the tower is β. Show that the height of the tower is h tan α cot β.

Let AB be the tower of height H and C be a point on the same horizontal level as B.

Let D be a point vertically above C such that CD = h.

Let BC = x.

From right angled ΔABC, we get

$\Rightarrow \cot \alpha = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] \Rightarrow \cot \alpha = \dfrac{x}{H} \\[1em] \Rightarrow x = H \cot \alpha \text{ …(1)}$

From right angled ΔDBC, we get

$\Rightarrow \cot \beta = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] \Rightarrow \cot \beta = \dfrac{x}{h} \\[1em] \Rightarrow x = h \cot \beta \text{ …(2)}$

From (1) and (2), we have

$\Rightarrow H \cot \alpha = h \cot \beta \\[1em] \Rightarrow H = h \cot \beta \times \dfrac{1}{\cot \alpha} \\[1em] \Rightarrow H = h \tan \alpha \cot \beta$

Hence, the height of the tower is h tan α cot β.