1. **Problem statement:** Given $m \angle JKM = 100^\circ$, $m \angle NKM = 29^\circ$, and $\overline{KN}$ bisects $\angle LKM$, find $m \angle JKL$. 2. **Understanding the problem:** The ray $KN$ bisects $\angle LKM$, so it divides $\angle LKM$ into two equal angles. Let $m \angle LKN = m \angle NKM = x$. 3. Since $m \angle NKM = 29^\circ$, then $m \angle LKN = 29^\circ$ as well. 4. Therefore, $m \angle LKM = m \angle LKN + m \angle NKM = 29^\circ + 29^\circ = 58^\circ$. 5. The angles around point $K$ on a straight line sum to $180^\circ$. Since $m \angle JKM = 100^\circ$ and $m \angle LKM = 58^\circ$, the remaining angle $m \angle JKL$ satisfies: $$m \angle JKM + m \angle JKL + m \angle LKM = 180^\circ$$ 6. Substitute known values: $$100^\circ + m \angle JKL + 58^\circ = 180^\circ$$ 7. Simplify: $$m \angle JKL = 180^\circ - 100^\circ - 58^\circ = 22^\circ$$ **Final answer:** $$m \angle JKL = 22^\circ$$