Let the distance between T 1 and T 2 be x From the figure EA = 60 m (T 1 ) and DB = 80 m (T 2 ) $\angle DEC = \theta$ 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 $\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}$ $\Rightarrow$ ${{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}$ $\Rightarrow$ ${x^2} = 1200$ $\Rightarrow$ $x = 20\sqrt 3$