A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$ before striking a point $B$ (which may be on $AD$ or $CD$) perpendicularly and retracing its path back to $A$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3$). If $\measuredangle CDA=8^\circ$, what is the largest value $n$ can have?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.}$
Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$, but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$, the reflection would give us a triangle with angles $16, 90$, and $74$. Then, when we reflect again, we will have $180 - 74 - 74$ = $32$. Since the other side of the reflection when we had the $82$ degrees had carried over to the other side, we have a $32-82-66$ triangle. 
Notice that we keep decreasing by increments of $8$. This is because the starting angle was $8$ and since we always have to decrease $8$ every time and that every triangle has every increasing angles of $8$, we must decrease by $8$ every time. This is the most optimal path of the light beam.
The pattern of light will be $82-74-66-58-50-42-34-26-18-10$. When we get to the angle of $2$ degrees, we have reached angle $A$. Therefore, we don't count the $2$, so our total number of reflections between $CD$ and $AD$ is $\boxed {(B) 10}$
~Arcticturn