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Two poles of equal lengths are standing opposite each other on either side of the roads,which is $80 m$ wide. From a point between them on the roads, the angle of elevation of the top of the poles are $60^{\circ}$ and $30^{\circ}$ respectively. Find the height of the poles and the distance of the point from the poles.

Class: 10

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60 m, 20 \sqrt{3} m

50 m, 20 \sqrt{3} m

60 m, 10 \sqrt{3} m

50 m, 10 \sqrt{3} m

answer is A.

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Detailed Solution

Given: AB and CD are the heights of the poles.
BD is the length of the road.
Let

AB

and

CD

be equal to

' h' m

.
Let

BO =' x' m

Then

OD = (80 - x) m

Δ AOB,

\Rightarrow \tan 30^{\circ} = \frac{AB}{OB}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x}

\Rightarrow h = \frac{x}{\sqrt{3}}

[Equation I] Question Image

Δ COD,

\Rightarrow \tan 60^{\circ} = \frac{CD}{OD}

\Rightarrow \sqrt{3} = \frac{h}{80 - x}

\Rightarrow h = (80 - x) \sqrt{3}

\Rightarrow \frac{x}{\sqrt{3}} = (80 - x) \sqrt{3}

[From equation I]

\Rightarrow x = (80 - x) \sqrt{3} \times \sqrt{3}

\Rightarrow x = (80 - x) \times 3

\Rightarrow x = 240 - 3 x

\Rightarrow x + 3 x = 240

\Rightarrow 4 x = 240

\Rightarrow x = \frac{240}{4}

\Rightarrow x = 60 m

And therefore,

h = \frac{x}{\sqrt{3}} = \frac{60}{\sqrt{3}} = \frac{60 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{60 \sqrt{3}}{3} = 20 \sqrt{3} m

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