 # ML Aggarwal Solutions Class 10 Mathematics Solutions for Heights and Distances Exercise 20 in Chapter 20 - Heights and Distances

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:

The angle produced by the line of sight and the horizontal plane for an object above the horizontal is known as the angle of elevation. As seen in the image below, we must take into consideration two right angled triangles to get the height of the lamp post.

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

In right △ AEC

tan θ = AE/CE

Substituting the values

tan θ = (x – 1.8)/ 3.6 …… (1)

In right △ CDF

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.

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