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° to 60° as he walks towards the building. Find the distance he walked towards the building.

To find: Distance travelled

Formula Used:

Explanation:

Let the initial point of boy is L and final point is M and x is the distance he walked.

QC = QP – CP

QC = 30 - 1.5

= 28.5 m.

In ∆QCA,

⇒ AC = 28.5√3 …. (1)

In ∆QCB,

AB = AC – BC

= 19√3 m

Therefore, the walking distance of boy is 19√3 m.