If we consider the rope to be taut, we can have a good approximation of a side view of the tent as follows:
If you consider the central triangle to be the tent, and the big black line the rope, we have the case of a right triangle, and we can use Pythagoras' theorem to solve it as follows:
[tex]\begin{gathered} h^2+x^2=l^2 \\ 4^2+x^2=8^2\to16+x^2=64 \\ x^2=64-16=48\to x\cong6.928\ldots \\ \end{gathered}[/tex]
making an approximation as it follows:
[tex]x\cong6.9ft[/tex]
