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.
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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