How far would Rebecca walk if she walks 100 feet in a straight line and turns $165^{\circ}$ clockwise, repeating this pattern until... If Rebecca walks 100 feet in a straight line. She turns $165^{\circ}$ clockwise and walks another 100 feet, and turns $165^{\circ}$ clockwise again. If she continues this pattern until she reaches the point where she started. How far did she walk?
I can't see how she would ever get back to her origin.
She turns $\frac{11}{24}$ of full rotation each time, this will not form a regular polygon.
It only makes sense she gets back to start if she turns for example $20^{\circ}$ - convenient factor of $360$, here I can use the fact the exterior angles of polygon add up $360^{\circ}$ to find the number of angles formed and derive an answer.
 A: The shape that Rebecca walks is an example of a
regular star polygon.
Unlike the sides of an ordinary regular polygon,
the sides of a regular star polygon intersect each other.
You can show that after walking a certain number of $100$-foot lengths and performing the $165$-degree turn at the end of each $100$ feet, Rebecca will have made several full turns plus an additional $180^\circ.$
That is, she will be facing exactly opposite the direction in which she first started walking. The next $100$ feet will cancel the first $100$ feet she walked, then she will cancel the next $100$ feet, and so forth until she has canceled all of the "certain number of" $100$-foot lengths and is back where she started.
I leave it as an exercise to figure out how many times Rebecca has to turn before she is facing opposite her original direction for the first time.
Note that the vertices of a regular star polygon are also vertices of an ordinary regular polygon. If you can figure out how many turns Rebecca makes to get back where she started, you can figure out the number of vertices of the star polygon.
You can then try to figure out how to inscribe the star polygon in a regular polygon with that number of vertices in case you are really curious to see what her path is like and exactly how she gets back to her starting point.
Technically, however, you don't need to draw the polygon in order to find the answer to the question.
You just need to know how many sides it has.

As a hint to how to figure out the number of vertices without having to use the internal-angle formula from the web page (which is suspect anyway, since it only accounts for the $p$ in the notation $\{p/q\}$, whereas $\{7/3\}$ clearly has a different angle than $\{7/2\}$),
consider that after turning $165^\circ$ clockwise at every vertex,
Rebecca must be facing in the same direction she started.
That is, she must have turned through a whole multiple of $360^\circ.$
Also consider that if there are $n$ vertices, Rebecca would have turned
$n \times 165^\circ$ clockwise
So try to see what is the least multiple of $165^\circ$ that is also a multiple of $360^\circ.$
As a further hint, notice that $165^\circ = 180^\circ - 15^\circ.$

A postscript:
I would also like to note that it is possible for a star polygon to have an odd number of sides (the five-pointed star is an obvious example), in which case none of the walks along the sides is canceled by another side in a $180$-degrees opposite direction. But the final displacement is the vector sum of all the displacements along the individual sides, which combine by adding the vectors; and the vector sum is zero. One way to see this is to rearrange the sequence of sides to that they form a regular convex polygon.
A: For visualization, Rebecca will walk the following path:

A: Here is an approach using the representation of points in the plane as complex numbers.
For convenience, let's pick units of measurement so that $1$ unit = $100$ feet, and let Rebecca's location at step $n$ be $z_n$ for $n = 0, 1, 2, \dots$, with initial location $z_0 = 0$ and position after the first step $z_1 = 1$.  Then her position after moving $1$ unit of distance after a turn of $-11/12 \pi$ radians is $z_2 = 1 + a$, where we define $a = \exp(-(11/12) \pi i)$.  Her position after another step is $z_3 = 1 + a + a^2$, and her position at the $n$th step is
$$z_n = 1 + a + a^2 + \dots a^{n-1}$$
Applying the formula for the sum of a geometric series,
$$z_n = \frac{1 - a^n}{1-a}$$
So Rebecca gets back to her initial location when $a^n = 1$, i.e.
$$\exp \left(- \frac{11}{12} n \pi i \right) = 1$$
Since $\exp(z) = 1$ exactly when $z = 2m \pi i$ for some integer $m$, we know
$$\frac{11}{12} n \pi i = 2 m \pi i$$
so we need the smallest positive integer $n$ such that
$$11n = 24 m$$
for some positive integer $m$.  I'll leave the final step to the reader.
