A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. Fromthe top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feetof the towers T1 and T2 is :
Solutions · 1
Official solution
Let the distance between T1 and T2 be x
From the figure
EA = 60 m (T1) and DB = 80 m (T2)
\[\angle DEC = \theta \] and \[\angle BEC = 2\theta \]
Now in \[\angle DEC\],
\[\tan \theta = {{DC} \over {AB}} = {{20} \over x}\]
and in \[\Delta BEC\],
\[\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}\]
We know that
\[\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}\]
\[ \Rightarrow \] \[{{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}\]
\[ \Rightarrow \] x2=1200 \[ \Rightarrow \] \[x = 20\sqrt 3 \]