From a window A, 10 m above the ground, the angle of elevation of the top C of a tower is x°, where tan x = $\Big(\dfrac{5}{2}\Big)$ and the angle of depression of the foot D of the tower is y°, where tan y = $\Big(\dfrac{1}{4}\Big)$.

See the figure given alongside. Calculate the height CD of the tower in metres.

⇒ AB = DE = 10 m.

In ∆AED,

$\Rightarrow \tan y^{\circ} = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan y^{\circ} = \dfrac{DE}{AE} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{DE}{AE} \\[1em] \Rightarrow AE = 4DE \\[1em] \Rightarrow AE = 4 \times 10 \\[1em] \Rightarrow AE = 40 \text{ m}.$

In ∆AEC,

$\Rightarrow \tan x^{\circ} = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan x^{\circ} = \dfrac{CE}{AE} \\[1em] \Rightarrow \dfrac{5}{2} = \dfrac{CE}{AE} \\[1em] \Rightarrow CE = \dfrac{5}{2}AE \\[1em] \Rightarrow CE = \dfrac{5}{2} \times 40 \\[1em] \Rightarrow CE = 100 \text{ m.}$

From figure,

CD = DE + CE = 10 + 100 = 110 m.

Hence, height of tower (CD) = 110 m.