Question:
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30 degree to 60 degree as he walks towards the building. Find the distance he walked towards the building.
Answer:
Let AE is the Length of the building.
So AE = 30
Again BE = DF = 1.5
AB = AE - BE
= 30 - 1.5
= 28.5
Now in triangle ABC,
tan60 = AB/BC
=> √3 = 28.5/BC
=> BC = 28.5/√3
Again in triangle ABD
tan30 = AB/BD
=> 1/√3 = 28.5/BD
=> BD = 28.5*√3
=> BC + CD = 28.5√3
=> 28.5/√3 + CD = 28.5√3
=> CD = 28.5√3 - 28.5/√3
=> CD = (28.5*3 - 28.5)/√3
=> CD = 28.5(3-1)/√3
=> CD = (28.5*2)/√3
=> CD = 57/√3
=> CD = 57√3/(√3*√3) (Multiply √3 in numerator and denominator)
=> CD = 57√3/3
=> CD = 19√3
So the distance he walked towords building = 19√3 m
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