A man 1.8 m high stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground. Find the height of the lamp post.

Let AB be the lamp post and CD the height of man.

BD is the distance of man from the foot of the lamp and FD is the shadow of man.

CE || DB.

Take AB = x and CD = 1.8 m
EB = CD = 1.8 m
CE = DB = 3.6 m
AE = (x - 1.8) m
Shadow (FD) = 5.4 m

Considering right angled △ACE, we get

$\Rightarrow \text{tan θ} = \dfrac{AE}{CE} \\[1em] \Rightarrow \text{tan θ} = \dfrac{x - 1.8}{3.6} \text{ ……(Eq 1)}$

Considering right angled △CFD, we get

$\Rightarrow \text{tan θ} = \dfrac{CD}{FD} \\[1em] \Rightarrow \text{tan θ} = \dfrac{1.8}{5.4} = \dfrac{1}{3} \text{ ……(Eq 2)}$

Comparing Eq 1 and Eq 2 we get,

$\Rightarrow \dfrac{x - 1.8}{3.6} = \dfrac{1}{3} \\[1em] \Rightarrow 3x - 5.4 = 3.6 \\[1em] \Rightarrow 3x = 5.4 + 3.6 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3.$

Hence, the height of the lamp post is 3 meters.