Question

A $1.5m$ tall boy is standing at some distance from a $30m$ tall building. The angle of elevation from his eyes to the top of the building increases from ${30}^{\circ }and{60}^{\circ }$ as he walks towards the building. Find the distance he walked towards the building.

Answer 1

Let the boy initially stand at point $Y$ with inclination $30°$ and then he approaches the building to the point $X$ with inclination $60°$.

To Find: The distance boy walked towards the building i.e. $XY$

From the diagram,, $XY=CD$.

Height of the building $=AZ=30m$.

$AB=AZ–BZ=30–1.5=28.5$

Measure of $AB$ is $28.5m$

In right $\Delta ABD$,

$tan30°=\frac{AB}{BD}\Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{BD}$

$BD=28.5\sqrt{3}m$

Also, In right $\Delta ABC$,

$tan60°=\frac{AB}{BC}\Rightarrow \sqrt{3}=\frac{28.5}{BC}$

$\Rightarrow BC=\frac{28.5}{\sqrt{3}}=\frac{28.5\sqrt{3}}{3}$

So, the length of $BC$ is $\frac{28.5\sqrt{3}}{3}m.$

$XY=CD=BD–BC=(28.5\sqrt{3}-\frac{28.5\sqrt{3}}{3}$

$=28.5\sqrt{3}(1-\frac{1}{3})$

$=28.5\sqrt{3}\times \frac{2}{3}$

$=\frac{57}{\sqrt{3}}m.$

Thus, the distance boy walked towards the building is $\frac{57}{\sqrt{3}}m.$