question_answer

Two poles of equal heights are standing opposite to each other on either side of a road which is 100 m wide from a point between them on road. Angles of elevation of their tops are \[30{}^\circ \]and \[60{}^\circ \]The height of each pole in metre) is           [SSC (10+2) 2011]

A) \[25\sqrt{3}\]    

B) \[20\sqrt{3}\]

C) \[28\sqrt{3}\]                

D) \[30\sqrt{3}\]

Correct Answer: A

Solution :

Let the height of two poles be h m each.

Given, distance between two poles \[=100\,\,m\]

Let the distance of first pole from the point \[=x\,\,m\]

Then, the distance of second pole from the point

\[=(100-x)\,\,m\]

In \[\Delta ABO,\]           \[\tan 30{}^\circ =\frac{h}{x}\]

\[\Rightarrow \]   \[\frac{1}{\sqrt{3}}=\frac{h}{x}=\sqrt{3}h=x\]              (i)

From \[\Delta DOC,\]\[\tan \,\,60{}^\circ =\frac{h}{100-x}\]

\[\Rightarrow \]               \[\sqrt{3}=\frac{h}{100-x}\]

\[\Rightarrow \]   \[\sqrt{3}\,\,(100-x)=h\]

\[\Rightarrow \]   \[\sqrt{3}\,\,(100-\sqrt{3}h)=h\]   [from Eq. (i)]

\[\Rightarrow \]   \[100\sqrt{3}-3h=h\]

\[\Rightarrow \]   \[4h=100\sqrt{3}\]\[\Rightarrow \]\[h=25\sqrt{3}\,\,m\]