The following diagonal is drawn in a regular heptagon, creating a pentagon and a quadrilateral.  What is the measure of $x$, in degrees?  Express your answer as a common fraction.

[asy]
import markers;
for(int i=0; i <=7; ++i) {
draw(dir(360*i/7+90)--dir(360*(i+1)/7+90));
}
pair A = dir(360*0/7+90);
pair F = dir(360*4/7+90);
pair G = dir(360*5/7+90);
pair H = dir(360*6/7+90);

draw(A--F);

markangle(Label("$x$",Relative(0.5)),n=1,radius=18,G,F,A);

[/asy]
The angles of a regular $n$-gon have measure $\left(\frac{180(n-2)}n\right)^\circ$.  Therefore the angles in a regular heptagon measure \[y=\frac{180\cdot5}7=\frac{900}7\]degrees.

We also note that since the larger angles of the quadrilateral are equal, and the three corresponding sides are equal, this is an isosceles trapezoid.  Therefore we get the following angles:

[asy]
import markers;
for(int i=0; i <=7; ++i) {
draw(dir(360*i/7+90)--dir(360*(i+1)/7+90));
}
pair A = dir(360*0/7+90);
pair F = dir(360*4/7+90);
pair G = dir(360*5/7+90);
pair H = dir(360*6/7+90);

draw(A--F);

markangle(Label("$x$",Relative(0.5)),n=1,radius=18,G,F,A);
markangle(Label("$x$",Relative(0.5)),n=1,radius=18,F,A,H);
markangle(Label("$y$",Relative(0.5)),n=1,radius=14,A,H,G);
markangle(Label("$y$",Relative(0.5)),n=1,radius=14,H,G,F);

[/asy]

The sum of the angle measures in a quadrilateral is always $360^\circ$, so we have  \[360=x+x+y+y=x+x+\frac{900}7+\frac{900}7.\]Therefore  \begin{align*}
x+x&=\frac{360\cdot7-900-900}7\\
&=\frac{180\cdot14-180\cdot5-180\cdot5}7\\
&=\frac{180\cdot4}7\\
&=\frac{720}7.\\
\end{align*}Since $x$ is half of that, $x=\boxed{\frac{360}7}$ degrees.