Question

A boy of height 150 cm tall is standing at some distance
from a 30 m tall building. The angle of elevation
increases from 30° to 60° as he walks towards the
building. Find the distance he walked towards the
building. [Use 3 =1.731​

Answer 1

★$\huge{\underline{\underline{\color{darkblue}{\bf{Given}}}}}$★

➜AB is the boy

➜AB = 150cm = 1.5m

➜ED is the building

➜ED = 30m

➜/_EFP = 60°

➜/_EAQ = 30°

➜EP = ED-PD = 30-1.5 = 28.5m

★$\huge{\underline{\underline{\color{darkblue}{\bf{To\: find}}}}}$★

➢The distance that boy walked towards building

AQ = BC = ?

★$\huge{\underline{\underline{\color{darkblue}{\bf{Solution}}}}}$★

➢In triangle EAP

${⇒tan30\degree}{=}\frac{EP}{AP}$

⇒1/√3 = $\frac{28.5}{AP}$

${⇒AP=28.5}\sqrt{3}$cm

━━━━━━━━━━━━━━━━━━━━━━━━

➢In triangle EQP

${⇒tan60\degree}{=}\frac{EP}{QP}$

⇒$\sqrt{3}{=}\frac{28.5}{QP}$

${⇒QP}{=}$ 28.5/√3cm

━━━━━━━━━━━━━━━━━━━━━━━━━

➢Now,

AQ = AP-QP

⇒AQ = 28.5√3-28.5/√3

⇒AQ = 28.5×3-28.5/√3

⇒AQ = 85.5 - 28.5 /√3

⇒AQ = 57/√3

⇒AQ = √3×√3×19/√3

⇒AQ = 19√3

⇒AQ = 19×1.73 = 32.87

$\huge{\boxed{\bf{\orange{AQ\:=\: 32.87m}}}}$

The distance that boy walked towards

building is 32.87m

Answer 2

$\Large\red{\underline{\underline{\bf{Given}}}}$

A boy of height 150 cm tall is standing at some distance from a 30 m tall building. The angle of elevation increases from 30° to 60° as he walks towards the building.

$\Large\red{\underline{\underline{\bf{Find\;out}}}}$

Find the distance he walked towards the building.

$\Large\red{\underline{\underline{\bf{Solution}}}}$

• FA = DC = 150cm = 1.5m
• FA = DC = 150cm = 1.5mGD = GC - CD = 30 - 1.5 = 28.5m

In ∆GFD

$\implies\sf tan30\degree=\Large\frac{GD}{FD}$

$\implies\sf \Large\frac{1}{\sqrt{3}}=\Large\frac{28.5}{FD}$

$\implies\sf FD=28.5\sqrt{3}m$

In ∆GED

$\implies\sf tan60\degree=\Large\frac{GD}{ED}$

$\implies\sf \sqrt{3}=\Large\frac{28.5}{ED}$

$\implies\sf \sqrt{3}ED=\Large\frac{28.5}{\sqrt{3}}m$

Distance he walked towards the building

=> FE = FD - ED

$\implies\sf FE=28.5\sqrt{3}-\Large\frac{28.5}{\sqrt{3}}$

$\implies\sf FE=\Large\frac{28.5×3-28.5}{\sqrt{3}}$

$\implies\sf FE=\Large\frac{85.5-28.5}{\sqrt{3}}$

$\implies\sf FE=\Large\frac{57}{\sqrt{3}}=19\sqrt{3}m$

$\implies\sf FE=19\times{1.731}=32.88m$