Question

A 1.5m tall girl is standing at some distance from a 30m tall building. The angle of elevation from her eyes to the top of the building increases from 30° to 60° as she walks towards the building. Find the distance she walked towards the building.

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Answer 1

Answer:

Given :-

• A 1.5 m tall girl is standing at some distance from a 30 m tall building.
• The angle of elevation from her eyes to the top of the building increases from 30° to 60° as she walks towards the building.

To Find :-

• What is the distance she walked towards the building.

Solution :-

Let,

$\mapsto$ Height of the building = AB

Given :

• ∠ADF = 30°
• ∠AEF = 60°

Then,

$\implies \sf AF =\: AB - FB$

• AB = 30 m
• FB = 1.5 m

$\implies \sf AF =\: 30\: m -\: 1.5\: m$

$\implies \sf AF =\: \bigg(30 - \dfrac{15}{10}\bigg)\: m$

$\implies \sf AF =\: \bigg(\dfrac{300 - 15}{10}\bigg)\: m$

$\implies \sf AF =\: \bigg(\dfrac{285}{10}\bigg)\: m$

$\implies \sf\bold{\green{AF =\: 28.5\: m}}$

Now,

$\dashrightarrow \sf\bold{In\: \Delta AFE\: :-}$

$\implies \sf tan 60^{\circ} =\: \dfrac{AF}{EF}$

Here,

• AF = 28.5 m

$\implies \sf \sqrt{3} =\: \dfrac{28.5}{EF}\: \: \bigg\lgroup \bold{\pink{tan 60^{\circ} =\: \sqrt{3}}}\bigg \rgroup$

$\implies \sf EF =\: \dfrac{28.5}{\sqrt{3}}$

$\implies \sf\bold{\purple{EF =\: \dfrac{28.5}{\sqrt{3}}\: m}}$

Again,

$\dashrightarrow \sf\bold{In\: \Delta AFD\: :-}$

$\implies \sf tan 30^{\circ} =\: \dfrac{AF}{DF}$

Here,

• AF = 28.5 m

$\implies \sf \dfrac{1}{\sqrt{3}} =\: \dfrac{28.5}{DF}\: \: \bigg\lgroup \bold{\pink{tan 30^{\circ} =\: \dfrac{1}{\sqrt{3}}}}\bigg \rgroup$

By doing cross multiplication we get,

$\implies \sf DF =\: 28.5\sqrt{3}$

$\implies \sf\bold{\purple{DF =\: 28.5\sqrt{3}\: m}}$

Now, we have to find the distance she walked towards the building :

$\leadsto \sf \bold{DE =\: DF - EF}$

Here we get,

• DF = 28.5√3 m
• EF = 28.5/√3

$\leadsto \sf DE =\: 28.5\sqrt{3} -\dfrac{28.5}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{28.5 \times 3 - 28.5}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{28.5(3 - 1)}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{28.5(2)}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{28.5 \times 2}{\sqrt{3}}$

$\leadsto \sf DF =\: \dfrac{\dfrac{285}{\cancel{10}} \times {\cancel{2}}}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{\dfrac{\cancel{285}}{\cancel{5}}}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{57}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{57}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{57 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$\leadsto \sf DE =\: \dfrac{57\sqrt{3}}{3}\: \bigg\lgroup \bold{\sqrt{3} \times \sqrt{3} =\: 3}\bigg\rgroup$

$\leadsto \sf DE =\: \dfrac{\cancel{57}\sqrt{3}}{\cancel{3}}$

$\leadsto \sf DE =\: \dfrac{19\sqrt{3}}{1}$

$\leadsto \sf\bold{\red{DE =\: 19\sqrt{3}\: m}}$

$\therefore$ The distance she walked towards the building is 19√3 m .

[ Note : Please refer the attachment for the diagram. ]

Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt A \:1.5\:m\: tall\: girl\: is\: standing\: at\: some\: distance\: from\: a\\\tt 30\:m \:tall\: building.\: The \:angle\: of\: elevation\: from\: her\\\tt eyes\: to \:the \:top of\: the\: building\: increases\: from\: {30}^{\circ}\\\tt to\: {60}^{\circ}\: as\: she\: walks\: towards\: the\: building. \:Find\: the\\\tt distance\: she \:walked \:towards \:the \:building.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\bf Heigh\:of \:the \:girl=1.5\:m$
• $\bf Height\:of \:the \:building =30\:m$
• $\bf Angle\: of\: elevation\: increased ={30}^{\circ}\:to\:{60}^{\circ}$

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

• $\bf Distance\: walked \:by\:the\:girl\:towards \:the \:building$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

${\pink {\underline {\bf {\pmb {Distance\: walked \:by\:the\:girl\:towards \:the \:building}}}}}$

${\orange {\mathfrak {Let,}}}$

${\blue {\tt {\: \: \: \: \: \: \:AD\: represents\: the\: distance \:walked\: by\: the\: girl}}}$

$\bf\leadsto BC=CG-BG$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies BG=30-1.5$

$\implies {\blue {\boxed {\boxed {\purple {\sf {BG=28.5}}}}}}$

$\bf\leadsto In\:\triangle ABC,\angle A={30}^{\circ}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {tan\:\theta=\dfrac{opposite}{adjacent}}}}}}}}$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies tan\:{30}^{\circ}=\dfrac{CB}{AB}$

$\bf \implies \dfrac{1}{\sqrt{3}}=\dfrac{28.5}{AB}$

$\implies {\blue {\boxed {\boxed {\purple {\sf {AB=28.5\sqrt{3}}}}}}}$

$\bf\leadsto In\:\triangle BCD,\angle D={60}^{\circ}$

${\blue {\boxed {\boxed {\boxed {\green {\pmb {tan\:\theta=\dfrac{opposite}{adjacent}}}}}}}}$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies tan\:{60}^{\circ}=\dfrac{CB}{BD}$

$\bf \implies \sqrt{3}=\dfrac{28.5}{BD}$

$\bf \implies BD\sqrt{3}=28.5$

$\implies {\blue {\boxed {\boxed {\purple {\sf {BD=\dfrac{28.5}{\sqrt{3}}}}}}}}$

$\bf\leadsto\:\: \therefore AD=AB-BD$

${\underbrace {\overbrace {\orange {\pmb {Substitute \:the \:values}}}}}$

$\bf \implies AD=28.5\sqrt{3}-\dfrac{28.5}{\sqrt{3}}$

$\bf \implies AD=\dfrac{28.5\times 3-28.5}{\sqrt{3}}$

$\bf \implies AD=\dfrac{85.5-28.5}{\sqrt{3}}$

$\bf \implies AD=\dfrac{57}{\sqrt{3}}$

$\bf \implies AD=\dfrac{57}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\bf \implies AD=\dfrac{57\sqrt{3}}{3}$

$\bf \implies AD=\dfrac{\cancel{57}\sqrt{3}}{\cancel{3}}$

$\implies {\blue {\boxed {\boxed {\purple {\mathfrak {AD=19\sqrt{3}\:m}}}}}}$

${\underbrace {\red {\underline {\red {\overline {\red {\pmb {\sf {{\therefore} The\:distance \:walked\:by\:the\:girl\:towards \:the \:building\:is\:19\sqrt{3}\:m}}}}}}}}}$

____________________________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

${\sf {\pink {Trigonometric \:ratios\:table}}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\theta & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \bf sin \:\theta & \sf 0 & \sf\dfrac{1}{2}& \sf\dfrac{1}{ \sqrt{2}} & \sf\dfrac{\sqrt{3}}{2} &\sf1 \\ \\ \bf cos \:\theta & \sf 1 &\sf \dfrac{\sqrt{3} }{2}&\sf \dfrac{1}{\sqrt{2}} &\sf \dfrac{1}{2} &\sf 0 \\ \\ \bf tan \:\theta &\sf 0 & \sf\dfrac{1}{\sqrt{3}}&\sf 1 & \sf\sqrt{3} & \sf \infty\end{array}}}\end{gathered}\end{gathered} \end{gathered}$