Answer
$\dfrac{\pi}{6}$
Work Step by Step
$\dfrac{17\pi}{6}$ is close to $\dfrac{18\pi}{6}=3\pi$.
This means that this angle is between $2.5\pi$ and $3\pi$.
RECALL:
An angle $\theta$, where $2.5\pi \lt \theta \lt 3\pi$, is coterminal with:
$\theta - 2\pi$
Thus, the given angle is coterminal with:
$=\dfrac{17\pi}{6} - 2\pi = \dfrac{17\pi}{6} - \dfrac{12\pi}{6} = \dfrac{5\pi}{6}$
$\dfrac{17\pi}{6}$ is coterminal with $\dfrac{5\pi}{6}$.
$\dfrac{5\pi}{6}$ is in Quadrant II so $\dfrac{17\pi}{6}$ is also in Quadrant II.
RECALL:
The following are the means on how to find the reference angle of an angle $0 \leq \theta \lt 2\pi$ based on its position:
(1) Quadrant I: $\theta$ (itself)
(2) Quadrant II: $\pi-\theta$
(3) Quadrant III: $\theta - \pi$
(4) Quadrant IV: $2\pi - \theta$
Use formula (2) above to obtain:
reference angle of $\dfrac{17\pi}{6}$ = reference angle of $\dfrac{5\pi}{6}$, which is
$\\=\pi - \dfrac{5\pi}{6}
\\=\dfrac{6\pi}{6} - \dfrac{5\pi}{6}
\\=\dfrac{\pi}{6}$
