$ABC$  is an equilateral triangle such that the vertices B and C lie on two parallel lines at a distance 6. If A lies between the parallel lines at a distance 4 from one of them then the length of a side of the equilateral triangle is 

From the choice of the axis  $A=(0,0),B=(2\cot \theta ,2),C=(4\cot ({60}^0-\theta ),-4)$
    
Now  ${\left(sideofequilateraltraingle\right)}^2$
 $$\begin{gathered} =4{\cot }^2\theta +4=16{\cot }^2(60-\theta )+16 \\ \Rightarrow 4\cos e{c}^2\theta =16\cos e{c}^2(60-\theta ) \\ \Rightarrow \tan \theta =\frac{\sqrt{3}}{5} \end{gathered}$$
 Hence the required length  $=2\cos ec\theta =4\sqrt{\frac{7}{3}}$