I tutored a student today and we were both stumped on the following question:
My thought process was that since $\angle$J is an inscribed angle, we can say that arc ILK = $222^{\circ}$ I don't think that gets me anywhere as I am not sure how to work with the lines $y - 2$ and $6$. Any hints would be greatly appreciated.
P.S. I know taking snapshots makes it rather difficult to track the problem if others had questions about it, but it came from my student's school portal, so I do not have direct access to its location.
Since $\;I\hat{J}K\;$ and $\;I\hat{L}K\;$ are opposite angles in a cyclic quadrilateral, their sum is $\;180^\text{o}$, hence
$I\hat{L}K=180^\text{o}-I\hat{J}K=180^\text{o}-111^\text{o}=69^\text{o}.$
Moreover
$y-2=6$
because tangent segments to a circle from the same external point $\;M\;$ are congruent.
Therefore
$y=8\;.$