Question

A 1.5 meter tall boy is standing at some distance from a 30 m tall building . The angle of elavation from his eyes to top of building rises from 30 to 60 degree as he walks towards building. Find the distancevhe walked

Answer 1

$\underline{\bold{A \: 1.5 \: meter \: tall \: boy \: is \: standing \: at \: some}}$ $\underline{\bold{distance \: from \: a \: 30 \: m \: tall \: building. \: The}}$ $\underline{\bold{angle \: of \: elevation \: from \: his \: eyes \: to \: top}}$ $\underline{\bold{of \: building \: rises \: from \: 30 \: to \: 60 \: degree \: as}}$ $\underline{\bold{he \: walks \: towards \: building. \: Find \: the}}$ $\underline{\bold{distance \: he \: walked.}}$

$\textbf {Let BC = FG = x and DC = EF = y}$

$\boxed {A.T.Q}$

Height of boy = DE = 1.5 m

Height of building = AG = 30 m

AB = AG - DE

= (30 - 1.5) m

= 28.5 m

Now

In ∆ABC

tan 60° = $\dfrac{AB}{BC}$

$\sqrt{3}$ = $\dfrac{28.5}{x}$

x = $\dfrac{28.5}{ \sqrt{3} }$ ____(1)

In ∆ABD

tan 30° = $\dfrac{AB}{BD}$

$\dfrac{1}{ \sqrt{3} }$ = $\dfrac{AB}{BC \: + \: CD}$

$\dfrac{1}{ \sqrt{3} }$ = $\dfrac{28.5}{x \: + \: y}$

On cross multiplying them we get;

x + y = $\sqrt{3} \: (28.5)$

$\dfrac{28.5}{ \sqrt{3} }$ + y = $28.5\sqrt{3}$

$\dfrac{28.5 \: + \: y \sqrt{3} }{ \sqrt{3} }$ = $28.5\sqrt{3}$

$28.5 \sqrt{3}$ + $y \sqrt{3}$ = 85.5

$y \sqrt{3}$ = 85.5 - 28.5

$y \sqrt{3}$ = 57

y = $\dfrac{57}{ \sqrt{3} } \: \times \: \dfrac{ \sqrt{3} }{ \sqrt{3} }$

y = $\dfrac{ 57\sqrt{3} }{3}$

y = $19 \sqrt{3}$ m

The distance walked by boy = y

= $19 \sqrt{3}$ m