Ex.9.1 Q6 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

A \(1.5\,\rm{m}\) tall boy is standing at some distance from a \(30\,\rm{m}\) tall building. The angle of elevation from his eyes to the top of the building increases from \(30^\circ\) to \(60^\circ\) as he walks towards the building. Find the distance he walked towards the building.

Text Solution

What is Known?

(i) Height of the boy \(= 1.5\,{\rm{ m}}\)

(ii) Height of the building \(= 30\,{\rm {m}}\)

(iii) Angle of elevation from his eyes to the top of the building increases from \(30^\circ\) to \(60^\circ\) as he walks toward the building.

What is Unknown?

Distance, the boy walked towards the building

Reasoning:

Trigonometric ratio involving (\(AP, PR\) and \( \angle R\)) and (\(AP, PQ\)  and \(\angle Q\)) is \(tan \,\theta\)

[Refer thet  diagram to visualise \(AP,  PR\) and \(PQ \)]

Distance walked towards the building \(RQ\) \(= PR – PQ\)

Steps:

In \(\Delta \mathrm{APR}\)

\[\begin{align} \tan R &=\frac{A P}{P R} \\ \tan 30^{\circ} &=\frac{28.5}{P R} \\ \frac{1}{\sqrt{3}} &=\frac{28.5}{P R} \\ P R &=28.5 \times \sqrt{3} \mathrm{m} \end{align}\]

In \(\Delta \mathrm{APQ}\)

\[\begin{align}{\tan \mathrm{Q}}&={\frac{A P}{P Q}} \\ {\tan 60^{\circ}}&={\frac{28.5}{P Q}}\\ \sqrt{3} &=\frac{28.5}{P Q} \\ P Q &=\frac{28.5}{\sqrt{3}} \mathrm{m} \end{align}\]

Therefore,

\[\begin{align}PR - PQ &= 28.5\sqrt 3  - \frac{{28.5}}{{\sqrt 3 }}\\ &= 28.5\left( {\sqrt 3  - \frac{1}{{\sqrt 3 }}} \right)\\&= 28.5\left( {\frac{{3 - 1}}{{\sqrt 3 }}} \right)\\&= 28.5\left( {\frac{2}{{\sqrt 3 }}} \right)\\&= \frac{{57}}{{\sqrt 3 }}\\&= \frac{{57}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}\\\rm&= \frac{{57\sqrt 3 }}{3}\\&= 19\sqrt 3 {\rm{m}}\end{align}\]

Distance the boy walked towards the building is \(19 \sqrt{3} \mathrm{m}\).