Measuring an equilateral triangle

Suppose that $A$ is a distance $4$ from the line $L$ meeting the vertex $B$ . We can assume that $A,B,C$ have coordinates $(\sqrt{D^2-16},4)$ , $(0,0)$ and $(X,6)$ , choosing coordinates so that $L$ lies along the $x$ -axis. Here $D$ is the side of the triangle. Then we require $D^2 \; = \; X^2 + 36 \; = \; (\sqrt{D^2-16}- X)^2 + 4$ and hence $2X\sqrt{D^2-16} \; =\; D^2 - 48$ Since $X^2 = D^2 - 36$ we deduce that $\begin{aligned} D^2 - 48 & = \; \pm2\sqrt{(D^2-16)(D^2-36)} \\ (D^2 - 48)^2 & = \; 4(D^2 - 16)(D^2 - 36) \\ 3D^4 - 112D^2 & = \; 0 \end{aligned}$ and hence $D = 4\sqrt{\tfrac73} = \boxed{6.110100927}$ . It turns out that $X < 0$ .

Let $\overline {BK}$ and $\overline {CL}$ be the parallel line segments through $B$ and $C$ respectively, and $\angle {ACL}$ be $α$ . Then $a\cos {(30\degree -α)}=6$ and $a\sin {α}=2$ , where $a$ is the length of each side of the triangle. Solving these we get $\cot {α}=\dfrac{5}{\sqrt {3}}, a=2\cosec {α}=4\sqrt \dfrac {7}{3}=\boxed {6.110}$ (approx)