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

Question 20 Heights and Distances Exercise 20

A man observes the angles of elevation of the top of a building to be 300. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 600. Find the height of the building correct to the nearest decimal place.

Answer:

Solution:

A commonly used concept in relation to height and distance is the angle of elevation, particularly in trigonometry. It is described as a relationship between an oblique line from the observer's eye to an item above his eye and the horizontal plane.

It is given that

AB is a building whose height be x

CD = 60 m

In △ ABC

tan 600 = x/BC

√3 = x/y

So we get

y = x/√3 ….. (1)

In △ ABD

tan 300 = AB/BD

1/√3 = x/ (y+ 60)

By cross multiplication

y + 60 = √3 x

y = √3 x – 60

Using both the equations we get

x/√3 = √3 x – 60

By further calculation

x = 3x – 60√3

3x – x = 60 × 1.732

So we get

x = (60 × 1.732)/ 2

x= height of building AB = 51.96 m

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