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 m towards the tower, the tangent of the angle is found to be $\dfrac{3}{4}$. Find the height of the tower.

Let the height of tower QR be h meters and the angle of elevation be θ₁ and θ₂ at points P and S respectively.

So,

$\text{tan }θ$1 = \dfrac{5}{12} \text{ and } \text{tan }θ2 = \dfrac{3}{4}

From figure,

QR = h meters
PQ = PS + QS = (192 + QS) meters.

Considering right angled △PQR, we get

$\Rightarrow \text{tan }θ_1 = \dfrac{QR}{PQ} \\[1em] \Rightarrow \dfrac{5}{12} = \dfrac{h}{192 + QS} \\[1em] \Rightarrow 5(192 + QS) = 12h \\[1em] \Rightarrow 5QS + 960 = 12h \text{ …….(Eq 1)}$

Considering right angled △SQR, we get

$\Rightarrow \text{tan }θ_2 = \dfrac{QR}{QS} \\[1em] \Rightarrow \dfrac{3}{4} = \dfrac{h}{QS} \\[1em] \Rightarrow 3QS = 4h \\[1em] \Rightarrow QS = \dfrac{4h}{3}\text{ …….(Eq 2)}$

Putting value of QS from Eq 2 in Eq 1 we get,

$\Rightarrow 5 \times \dfrac{4h}{3} + 960 = 12h \\[1em] \Rightarrow \dfrac{20h}{3} + 960 = 12h \\[1em] \Rightarrow \dfrac{20h + 2880}{3} = 12h \\[1em] \Rightarrow 20h + 2880 = 36h \\[1em] \Rightarrow 36h - 20h = 2880 \\[1em] \Rightarrow 16h = 2880 \\[1em] \Rightarrow h = \dfrac{2880}{16} \\[1em] \Rightarrow h = 180.$

Hence, the height of the tower is 180 meters.