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 *. 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 30o = PR/AR
1/ √3 = 28.5/ AR
AR = 28.5√3
In ΔPRB,
tan 60o = 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.
