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

Answer 1

Given,

The height of the tall boy (AS) = 1.5 m

The length of the building (PQ) = 30 m

Let the initial position of the boy be S. And, then he walks towards the building and reached at the point T.

From the fig. we have

AS = BT = RQ = 1.5 m

PR = PQ – RQ = (30 – 1.5)m = 28.5 m

In ΔPAR,

tan 30^o = PR/AR

1/ √3 = 28.5/ AR

AR = 28.5√3

In ΔPRB,

tan 60^o = PR/BR

√3 = 28.5/ BR

BR = 28.5/√3 = 9.5√3

So, ST = AB = AR – BR = 28.5√3 – 9.5√3 = 19√3

Therefore, the distance which the boy walked towards the building is 19√3 m.