Q11 of 61 Page 215

A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.


Given, AF = 30m, DF = BE = 1.5m,


AD = 28.5 m


In triangle ABD


ABD = 30°


We know,





BD = 28.5 √3


Now, in triangle ACD,


ABD = 60°






Multiplying and dividing the fraction by √3, we get



CD = 9.5√3


distance he walked towards the building = BD – CD


= 28.5√3 –9.5√3


= 19√3 m


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