Question

Height 1.5 meter boy standing at some distance from 30meter tall building the angle of elevation from his eyes to top of the building increases from 30° to 60° as he walks towards 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