Question

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The
elevation from his eyes to the top of the building increases from 30° to 60° ask
towards the building. Find the distance he walked towards the building.
hend the neceflavotion of the bottom​

Answer 1

Solution :

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

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The elevation from his eyes to the top of the building increases from 30° to 60° ask towards the building.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The distance he walked towards the building.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have;

• A boy stand DC = 1.5 m
• Tall Building AB = 30 m
• A0 = (AB-OB)=(30-1.5)m = 28.5 m
• Elevation ∠D= 30°
• Elevation ∠D' = 60°

$\underline{\underline{\bf{According\:to\:the\:question\::}}}}}$

$\dag$ In right angled Δ ADO :

$\longrightarrow\sf{tan\theta=\dfrac{Perpendicular}{Base} }\\\\\\\longrightarrow\sf{tan\theta=\dfrac{AO}{OD} }\\\\\\\longrightarrow\sf{tan30\degree=\dfrac{28.5}{r+m} }\\\\\\\longrightarrow\sf{\dfrac{1}{\sqrt{3} } =\dfrac{28.5}{r+m} }\\\\\\\longrightarrow\sf{\blue{r+m=28.5\sqrt{3} ........................(1)}}$

$\dag$ In right angled Δ OAD' :

$\longrightarrow\sf{tan\theta=\dfrac{Perpendicular}{Base} }\\\\\\\longrightarrow\sf{tan\theta=\dfrac{AO}{D'O} }\\\\\\\longrightarrow\sf{60\degree=\dfrac{28.5}{m} }\\\\\\\longrightarrow\sf{\sqrt{3} =\dfrac{28.5}{m} }\\\\\\\longrightarrow\sf{\sqrt{3} m=28.5}\\\\\\\longrightarrow\sf{m=\dfrac{28.5}{\sqrt{3} }} \\\\\\\longrightarrow\sf{m=\dfrac{28.5\times \sqrt{3} }{\sqrt{3} \times \sqrt{3} }\:\:\:[rationalise] }\\\\\\\longrightarrow\sf{m=\dfrac{\cancel{28.5}\sqrt{3} }{\cancel{3}} }$

$\longrightarrow\sf{\blue{m=9.5\sqrt{3} \:m}}$

Putting the value of m in equation (1),we get;

$\longrightarrow\sf{r+9.5\sqrt{3} =28.5\sqrt{3} }\\\\\\\longrightarrow\sf{r=\big(28.5\sqrt{3} -9.5\sqrt{3} \big)m}\\\\\\\longrightarrow\sf{\blue{r=19\sqrt{3}\:m }}$

Thus;

$\dag\underbrace{\bf{The\:distance\:he\:walked\:the\:building\:is\:r=\boxed{\sf{19\sqrt{3}m}}}}}$

Answer 2

Answer:

$\implies$ The distance walking the building is r = 19√3.

$\large\underline\mathrm{Given:-}$

• A 1.5 m tall boy is standing at some distance from a 30 m tall building. The
• elevation from his eyes to the top of the building.

$\large\underline\mathrm{To \: find}$

• The distance he walked to wards the building.

$\large\underline\mathrm{Solution}$

We have.

$\implies$ DC = 1.5m

$\implies$ AC =30m

$\implies$ AC = (AB - OB) = (30 - 1.5) = 28.5m

$\implies$ ∆D = 30°

$\implies$ ∆D' = 60°

In right angle ∆ADO:

$\implies$ tan(theta) = p/b

$\implies$ tan(theta) = AO/OD

$\implies$ tan30° = 28.5/(r + m)

$\implies$ 1/√3 = 28.5/(r + m). .....(1)

In right angle ∆OAD

$\implies$ tan(theta) = p/b

$\implies$ tan(theta) = AO/D'O

$\implies$ tan 60° = 28.5/m

$\implies$ √3 = 28.5/m

$\implies$ m = 28.5/√3

$\implies$ m = 28.5/√3 × √3/√3

$\implies$ m = 28.5√3/3

$\implies$ m = 9.5√3m

Putting the value of m in eq. (1) we get.

$\implies$ r + 9.5√3 - 28.5√3

$\implies$ r + (28.5√3 - 9.5√3)m

$\implies$ r = 19√3m

The distance walking the building is r = 19√3.