Answer:
5√6 mi
Explanation:
In the given figure, triangles DAC and EBC are similar.
The ratio of corresponding sides are:
[tex]\frac{DA}{EB}=\frac{AC}{BC}=\frac{DC}{EC}[/tex]Substitute the given values:
[tex]\begin{gathered} \frac{4\sqrt[]{138}}{\sqrt[]{138}}=\frac{BC+6\sqrt[]{3}}{BC} \\ 4=\frac{BC+6\sqrt[]{3}}{BC} \\ 4BC=BC+6\sqrt[]{3} \\ 4BC-BC=6\sqrt[]{3} \\ 3BC=6\sqrt[]{3} \\ BC=\frac{6\sqrt[]{3}}{3} \\ BC=2\sqrt[]{3} \end{gathered}[/tex]Since we already have BC and EB, we use the Pythagoras theorem:
[tex]\begin{gathered} EC^2=EB^2+BC^2 \\ EC^2=(\sqrt[]{138})^2+(2\sqrt[]{3})^2 \\ EC^2=138+12 \\ EC^2=150 \\ EC^{}=\sqrt{150} \\ EC=5\sqrt{6}\text{ mi} \end{gathered}[/tex]The exact length of EC is 5√6 mi.