A $ 1.5 \mathrm{~m} $ tall boy is standing at some distance from a $ 30 \mathrm{~m} $ tall building. The angle of elevation from his eyes to the top of the building increases from $ 30^{\circ} $ to $ 60^{\circ} $ as he walks towards the building. Find the distance he walked towards the building.
Given:
A \( 1.5 \mathrm{~m} \) tall boy is standing at some distance from a \( 30 \mathrm{~m} \) tall building.
The angle of elevation from his eyes to the top of the building increases from \( 30^{\circ} \) to \( 60^{\circ} \) as he walks towards the building.
To do:
We have to find the distance he walked towards the building.
Solution:
Let $AB$ be the height of the boy and $CD$ be the height of the building.
From the figure,
$\mathrm{AB}=\mathrm{OP}=\mathrm{DE}=1.5 \mathrm{~m}, \angle \mathrm{CAE}=30^{\circ}, \angle \mathrm{COE}=60^{\circ}$
Let the distance he walked towards the building be $\mathrm{AO}=x \mathrm{~m}$.
This implies,
$\mathrm{CE}=30-1.5=28.5 \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { CE }}{OE}$
$\Rightarrow \tan 60^{\circ}=\frac{28.5}{OE}$
$\Rightarrow \sqrt3=\frac{28.5}{OE}$
$\Rightarrow OE=\frac{28.5}{\sqrt3} \mathrm{~m}$.........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { CE }}{AE}$
$\Rightarrow \tan 30^{\circ}=\frac{28.5}{x+OE}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{28.5}{x+OE}$
$\Rightarrow x+OE=28.5\sqrt3 \mathrm{~m}$
$\Rightarrow x=28.5\sqrt3-\frac{28.5}{\sqrt3} \mathrm{~m}$ [From (i)]
$\Rightarrow x=\frac{28.5\sqrt3(\sqrt3)-28.5}{\sqrt3} \mathrm{~m}$
$\Rightarrow x=\frac{28.5(3-1)}{\sqrt3}\times \frac{\sqrt3}{\sqrt3} \mathrm{~m}$
$\Rightarrow x=\frac{28.5(2)\times\sqrt3}{3} \mathrm{~m}$
$\Rightarrow x=9.5(2)\sqrt3=19\sqrt3 \mathrm{~m}$
Therefore, the distance he walked towards the building is $19\sqrt3 \mathrm{~m}$.
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