Q18 of 28 Page 93

A 1.5 m long boy is standing at a certain distance from a 30 m high building when he goes towards the high building then the angle of elevation of the top of the building from his eye becomes 60° from 30°. Find by how much distance he has walked towards the building.


Let Height of the boy is DE, Height of the Tower is GC.


Given DE = 1.5, GC = 30, ∠CDB = 30o∠CAB = 60o. BC = 28.5, GC = 30.


in CDB,




BD = 28.5√3


in CAB,






Distance Travelled by boy = BD – AB = 28.5– 9.5


= 19


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