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.
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.

Solution:

Consider

AB as the lamp post

CD is the height of man

BD is the distance of man from the foot of the lamp

FD is the shadow of man

Construct CE parallel to DB

Take AB = x and CD = 1.8 m

EB = CD = 1.8 m

AE = x – 1.8

Shadow FD = 5.4 m

In right triangle ACE

tan θ = AE/CE

Substituting the values

tan θ = (x – 1.8)/ 3.6 …… (1)

In right triangle CFD

tan θ = CD/FD

Substituting the values

tan θ = 1.8/5.4 = 1/3 ….. (2)

Using both the equations

(x – 1.8)/ 3.6 = 1/3

So we get

3x – 5.4 = 3.6

3x = 3.6 + 5.4 = 9.0

By division

x = 9/3 = 3.0

Hence, the height of lamp post is 3 m.