A 1.5 m tall boy is standing at some distance from a 30 m tall building. Find the distance he walked towards the building.

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

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m

AB = AZ - BZ = (30 - 1.5) = 28.5 m
A/q,

In  right ΔABD,

tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD

⇒ BD = 28.5√3 m
also,
In  right ΔABC,

tan 60° = AB/BC
⇒ √3 = 28.5/BC

⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.