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.

Let AB be the initial position of boy and CD be the tower.

From figure,

CG = CD - GD = 30 - 1.5 = 28.5 m

In △AGC,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CG}{AG} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{28.5}{AG} \\[1em] \Rightarrow AG = 28.5 \times \sqrt{3} \\[1em] \Rightarrow AG = 28.5\sqrt{3} \text{ m}.$

In △EGC,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{CG}{EG} \\[1em] \Rightarrow \sqrt{3} = \dfrac{28.5}{EG} \\[1em] \Rightarrow EG = \dfrac{28.5}{\sqrt{3}} \text{ m}.$

From figure,

$\Rightarrow AE = AG - EG \\[1em] = 28.5\sqrt{3} - \dfrac{28.5}{\sqrt{3}} \\[1em] = \dfrac{28.5 \times 3 - 28.5}{\sqrt{3}} \\[1em] = \dfrac{85.5 - 28.5}{\sqrt{3}} \\[1em] = \dfrac{57}{\sqrt{3}} \\[1em] = \dfrac{57}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] = \dfrac{57\sqrt{3}}{3} \\[1em] = 19\sqrt{3} \text{ m}.$

Hence, the boy walked $19\sqrt{3}$ m towards the building.