Ex.9.1 Q10 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

Two poles of equal heights are standing opposite each other on either side of the road, which is \(80\,\rm{m}\) wide. From a point between them on the road, the angles of elevation of the top of the poles are \(60^\circ\) and \(30^\circ\) respectively. Find the height of the poles and the distances of the point from the poles.

   

Text Solution

   

What is Known?

(i) The poles of equal height.

(ii) Distance between poles \(= 80\,\rm{m}\)

(iii) From a point between the poles, the angle of elevation of the top of the poles are \(60^\circ\) and \(30^\circ\) respectively.

What is Unknown?

Height of the poles and the distances of the point from the poles.

Reasoning:

Let us consider the two poles of equal heights as \(AB\) and \(DC\) and the distance between the poles as \(BC\). From a point \(O\), between the poles on the road, the angle of elevation of the top of the poles \(AB\) and \(CD\) are \(60^\circ\) and \(30^\circ\) respectively.

Trigonometric ratio involving angles, distance between poles and heights of poles is \(\tan \theta \)

Steps:

Let the height of the poles be \(x\)

Therefore \(AB=DC=x\)

In \( \Delta AOB,\)

\[\begin{align}\text{tan 6}{{\text{0}}^{0}}&=\frac{AB}{BO} \\ \sqrt{3}&=\frac{x}{BO} \\ BO&=\frac{x}{\sqrt{3}}\qquad\left( \text{i} \right)  \end{align}\]

In \( \Delta OCD,\)

\[\begin{align}& \tan \text{ }{{30}^{0}}=\frac{DC}{OC} \\& \frac{1}{\sqrt{3}}=\frac{x}{BC-BO} \\ 
 & \frac{1}{\sqrt{3}}=\frac{x}{80-\frac{x}{\sqrt{3}}}\qquad\left[ \text{from }\left( \text{i} \right) \right] \\ 
 & 80-\frac{x}{\sqrt{3}}=\sqrt{3}x \\ & \frac{x}{\sqrt{3}}+\sqrt{3}x=80 \\ & x\left( \frac{1}{\sqrt{3}}+\sqrt{3} \right)=80 \\ & x\left( \frac{1+3}{\sqrt{3}} \right)=80 \\ & x\left( \frac{4}{\sqrt{3}} \right)=80 \\ & x=\frac{80\sqrt{3}}{4} \\ & x=20\sqrt{3} \\ \end{align}\]

Height of the pole \(x = 20\sqrt 3 {\rm{ }}\,\rm m\)

Distance of point from \(O\) the poles \(AB\)

\[\begin{align} BO&=\frac{x}{\sqrt{3}} \\ & =\frac{20\sqrt{3}}{\sqrt{3}} \\ & =20  \end{align}\]

Distance of point from \(O\) the poles \(CD\)

\[\begin{align}OC&=BC-BO \\ & =80-20 \\ & =60  \end{align}\]

Height of the poles are  \(20\sqrt{3}\text{ }\rm{m}\)  and the distances of the point from the poles are \(20\rm{m}\text{ and }60\rm{m}\) .