Question

A . A 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.
​

Answer 1

Answer:

19√3

EXPLANATION:-

Let AB=CD

HENCE, =EQ = 1.5 m

Let dist. covered by boy is BD=x

SINCE BD=AC=x

xLet DQ=CE=y

In fig. PE=30-1.5=28.5m

Angles are 30° and 60°

In ∆ PAE

In ∆ PAEtan 30 = PE/AE1/√3=28.5/x+y

In ∆ PAEtan 30 = PE/AE1/√3=28.5/x+y28.5√3=x+y

THIS IS (1)

In ∆PCE

In ∆PCEtan60=PE/CE√3=28.5/y

In ∆PCEtan60=PE/CE√3=28.5/yy=28.5/√3

In ∆PCEtan60=PE/CE√3=28.5/yy=28.5/√3Put in (1)x+28.5/√3=28.5√4x=(28.5×3-28.5)/√3= (85.5-28.5)/√3=57/√3Rationalize=57/√3×√3/√3=57√3/3

√3=28.5√4x=(28.5×3-28.5)/√3= (85.5-28.5)/√3=57/√3Rationalize=57/√3×√3/√3=57√3/3=19√3

Answer 2

$\small\bold{\underline{\sf{\blue{solution:-}}}}$

Let AB=CD

HENCE, =EQ = 1.5 m

Let dist. covered by boy is BD=x

SINCE BD=AC=x

xLet DQ=CE=y

In fig. PE=30-1.5=28.5m

Angles are 30° and 60°

In ∆ PAE

In ∆ PAEtan 30 = PE/AE1/√3=28.5/x+y

In ∆ PAEtan 30 = PE/AE1/√3=28.5/x+y28.5√3=x+y

THIS IS (1)

In ∆PCE

In ∆PCEtan60=PE/CE√3=28.5/y

In ∆PCEtan60=PE/CE√3=28.5/yy=28.5/√3

In ∆PCEtan60=PE/CE√3=28.5/yy=28.5/√3Put in (1)x+28.5/√3=28.5√4x=(28.5×3-28.5)/√3= (85.5-28.5)/√3=57/√3Rationalize=57/√3×√3/√3=57√3/3

√3=28.5√4x=(28.5×3-28.5)/√3= (85.5-28.5)/√3=57/√3Rationalize=57/√3×√3/√3=57√3/3=19√3