Question

If 1.5meter tall man stands at a distance of 50m from lamp post at night and it is observed that his shadow is 5m . find the height of the lamp post

Answer 1

Let AB as the lamp post that is x m in height and perpendicular on the ground . DE = 1.5m is the height of the man who is also standing perpendicular on the ground . BE= 50m is the distance between the lamp post and the man . EC = 5m is the shadow of the man .

→ Now taking the ∆ ABC and ∆ DEC .

Here ,

→ Angle ABC = Angle DEC [ perpendicular to the ground i.e 90° ]

→ Angle BCA = Angle ECD [ Common angles] .

So, ∆ ABC ~ ∆ DEC .

It means that fraction of their corresponding sides is also Equal .

$\frac{ab}{de} = \frac{bc}{ec}$

BC = BE + EC

BC = 50 + 5

BC = 55m

$\frac{x}{1.5} = \frac{55}{5}$

$x = 11 \times 1.5$

x = 16.5 m

So, the height of lamp post is 16.5 m

$\huge\mathbb\red{Hope\:it\:helps}$

Answer 2

SOLUTION:-

Given:

If 1.5m tall man stands at a distance of 50m from lamp post at night & it's observed that his shadow is 5m.

To find:

The height of the lamp post.

Explanation:

•Let AB be the lamp post of height R m

•Let CD be the height of the man.

•Let DE be the length of shadow of the man.

We have,

• CD=1.5m
• AD= 50m
• DE= 5m

In right angled ∆CDE,

$tan \theta = \frac{Perpendicular}{Base}$

So,

$tan \theta = \frac{CD}{DE} \\ \\ tan \theta = \frac{1.5}{5} = \frac{1.5 \times 10}{5 \times 10} \\ \\ tan \theta = \frac{15}{50} = 0.3m...............(1)$

&

In right angled ∆BAE;

$tan \theta = \frac{ R }{AD + DE} \\ \\ 0.3 = \frac{R}{50m + 5m} \\ \\ R = (55 \times 0.3)m \\ \\ R = 16.5m$

Thus,

The height of the lamp post is 16.5m.