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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.

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sir this is the important  question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
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.

question no 20 , heights and distances , ICSE board

 

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  1. It is given that

    AB is a building

    CD = 60 m

    ML Aggarwal Solutions for Class 10 Chapter 20 Image 20

    In triangle ABC

    tan 600 = AB/BC

    √3 = AB/BC

    So we get

    BC = AB/√3 ….. (1)

    In triangle ABD

    tan 300 = AB/BD

    1/√3 = AB/ (BC + 60)

    By cross multiplication

    BC + 60 = √3 AB

    BC = √3 AB – 60

    Using both the equations we get

    AB/√3 = √3 AB – 60

    By further calculation

    AB = 3AB – 60√3

    3AB – AB = 60 × 1.732

    So we get

    AB = (60 × 1.732)/ 2

    AB = 51.96 m

     

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