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 1

let AB=CD= EQ = 1.5 m

Let dist. covered by boy is BD=x

Let BD=AC=x

Let DQ=CE=y

In fig. PE=30-1.5=28.5m

Angles are 30° and 60°

In ∆ PAE

tan 30 = PE/AE

1/√3=28.5/x+y

28.5√3=x+y –(1)

In ∆PCE

tan60=PE/CE

√3=28.5/y

y=28.5/√3

Put in (1)

x+28.5/√3=28.5√4

x=(28.5×3-28.5)/√3

= (85.5-28.5)/√3

=57/√3

Rationalize

=57/√3×√3/√3

=57√3/3

=19√3

I HOPE ITS HELPFUL