NCERT · Class 10 · Mathematics · Applications of TrigonometryA 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 30 m tall building and $CD = 1.5$ m be the boy. The height of the building above the boy's eyes is: $$AE = 30 - 1.5 = 28.5 \text{ m}$| \nLet the boy initially stand at point $C$ where the angle of elevation is 30°, and walk to point $F$ where the angle of elevation becomes 60°. \nIn right-angled triangle $\triangle AEF$: $$\tan(60°) = \frac{AE}{EF} \implies \sqrt{3} = \frac{28.5}{EF} \implies EF = \frac{28.5}{\sqrt{3}}$| \nIn right-angled triangle $\triangle AEC$: $$\tan(30°) = \frac{AE}{EC} \implies \frac{1}{\sqrt{3}} = \frac{28.5}{EC} \implies EC = 28.5\sqrt{3}$| \nThe distance walked by the boy is $CF = EC - EF$: $$CF = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = 28.5 \left( \frac{3 - 1}{\sqrt{3}} \right) = \frac{28.5 \times 2}{\sqrt{3}} = \frac{57}{\sqrt{3}}$| \nRationalizing: $$CF = \frac{57\sqrt{3}}{3} = 19\sqrt{3} \text{ m}$| \nThus, the distance walked towards the building is $19\sqrt{3}$ m.