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10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

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Given that,
The height of both poles are equal  DC = AB. The angle of elevation of of the top of the poles are \angle DEC=30^o and \angle AEB=60^o resp.
Let the height of the poles be  h m and CE = x and BE = 80 - x

According to question,
In triangle DEC,

\\\Rightarrow \tan 30^o = \frac{DC}{CE} = \frac{h}{x}\\\\\Rightarrow \frac{1}{\sqrt{3}}= \frac{h}{x}\\\\\Rightarrow x=h\sqrt{3}..............(i)

In triangle AEB,
\\\Rightarrow \tan 60^o = \frac{AB}{BE}=\frac{h}{80-x}\\\\\Rightarrow \sqrt{3}=\frac{h}{80-x}\\\\\Rightarrow x=80 - \frac{h}{\sqrt{3}}..................(ii)
On equating eq (i) and eq (ii), we get

\sqrt{3}h=80 - \frac{h}{\sqrt{3}}
\frac{h}{\sqrt{3}}=20
h=20\sqrt{3} m
So, x = 60 m

Hence the height of both poles is (h=20\sqrt{3})m and the position of the point is at 60 m from the pole CD and 20 m from the pole AB. 

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manish

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