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.

   

 Video Solution
Some Applications Of Trigonometry
Ex 9.1 | Question 10

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 \( \Delta AOB,\)

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

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 \(20\rm{m}\text{ and }60\rm{m}\) .