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A man observes the angles of elevation of the top of a building to be 30^{0}. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60^{0}. 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 60^{0} = x/BC

√3 = x/y

So we get

y = x/√3 ….. (1)

In △ ABD

tan 30^{0} = 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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