At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $\dfrac{5}{12}$. On walking 192 meters towards the tower; the tangent of the angle is found to be $\dfrac{3}{4}$. Find the height of the tower.

Let's assume AB to be the vertical tower and C and D be the two points such that CD = 192 m.

Let ∠ACB = θ and ∠ADB = α

Given,

$\Rightarrow \text{tan θ} = \dfrac{5}{12} \\[1em] \Rightarrow \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{5}{12} \\[1em] \Rightarrow \dfrac{AB}{BC} = \dfrac{5}{12} \\[1em] \Rightarrow AB = \dfrac{5}{12}BC ………(1)$

Also given,

$\Rightarrow \text{tan α} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{AB}{BD} = \dfrac{3}{4} \\[1em] \Rightarrow AB = \dfrac{3}{4}BD ………(2)$

From equations (1) and (2), we get :

$\Rightarrow \dfrac{5}{12}BC = \dfrac{3}{4}BD \\[1em] \Rightarrow \dfrac{BC}{BD} = \dfrac{3 \times 12}{4 \times 5} \\[1em] \Rightarrow \dfrac{BD + CD}{BD} = \dfrac{36}{20} \\[1em] \Rightarrow \dfrac{BD + CD}{BD} = \dfrac{9}{5} \\[1em] \Rightarrow 5(BD + CD) = 9BD \\[1em] \Rightarrow 5BD + 5CD = 9BD \\[1em] \Rightarrow 9BD - 5BD = 5CD \\[1em] \Rightarrow 4BD = 5CD \\[1em] \Rightarrow BD = \dfrac{5 \times 192}{4} \\[1em] \Rightarrow BD = \dfrac{960}{4} \\[1em] \Rightarrow BD = 240 \text{ m}.$

BC = BD + DC = 240 + 192 = 432 m.

From equation (1),

AB = $\dfrac{5}{12} \times BC = \dfrac{5}{12} \times 432$

= 5 × 36

= 180 m.

Hence, the height of the tower is 180 m.