At a point on a level ground, the angle of elevation of the top of a tower is θ such that tan θ = $\Big(\dfrac{7}{12}\Big)$. On walking 64 m towards the tower, the angle of elevation is φ, where tan φ = $\Big(\dfrac{3}{4}\Big)$. Find the height of the tower.

Let h be the height of the tower (AB), x be the distance from foot of tower to second observation D,

Since the man walked 64 m towards the tower, the distance from C to the tower is (x + 64) m.

In right angled triangle ABD,

$\Rightarrow \tan \phi = \dfrac{\text{perpendicular}}{\text{base}} = \dfrac{h}{x} \\[1em] \Rightarrow \dfrac{3}{4} = \dfrac{h}{x} \\[1em] \Rightarrow x = \dfrac{4h}{3}.$

In right angled triangle ABC,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} = \dfrac{h}{x + 64} \\[1em] \Rightarrow \dfrac{7}{12} = \dfrac{h}{x + 64} \\[1em] \Rightarrow \dfrac{7}{12} = \dfrac{h}{\Big(\dfrac{4h}{3}\Big) + 64} \\[1em] \Rightarrow 7 \times \Big(\dfrac{4h}{3} + 64\Big) = 12h \\[1em] \Rightarrow \Big(\dfrac{28h}{3} + 448\Big) = 12h \\[1em] \Rightarrow 28h+ 1344 = 36h \\[1em] \Rightarrow 1344 = 36h - 28h \\[1em] \Rightarrow 1344 = 8h \\[1em] \Rightarrow h = \dfrac{1344}{8} \\[1em] \Rightarrow h = 168 \text{ m}.$

Hence, height of the tower is 168 m.