**Answer**

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.**