Question

1.5 m tall boy standing at some distance from a 30 M tall building the angle of elevation from his at the top of building increases from 30 to 60 degrees he works toward the building find the distance he walked towards the building

Answer 1

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 towards building = 19√3 m