$|\angle CDL|=0.5(|arc(CL)|-|arc(AB)|)$
The point $D$ is exterior to the circle. The segments $\overline{DC}~\&~\overline{DL}$ both intersect the circle "cutting off" two arcs.
It is a theorem that $|\angle CDL|$ is one-half the difference in the measures of the major & minor arcs.