We have that each of $\angle CAK, \angle KAC$, and $\angle BAC$ are $20^{\circ}$. We also know that $ABLM$ is an isosceles trapezoid with where $\angle BCM$ is $120^{\circ}$. In $\triangle ABL$, we know that $\angle BLA$ is $100^{\circ}$ so then by subtraction, we know that $\angle ALM$ is $20^{\circ}$. Also, $\triangle CMK$ is a $30-60-90$ triangle where $\angle MKC$ is right. Thus we know that $\angle MKL$ is also right. Since $\triangle AKL$ is isosceles with $\angle KAL$ being $20^{\circ}$, $\angle AKL$ is $80^{\circ}$ and it follows that $\angle AKM=10^{\circ}$. Ergo, $ \angle AKM+\angle ALM= 10+20=\boxed{30^{\circ}}$

See attached; $\angle AKM=\angle BAK=\angle LAC$, hence $\angle AKM+\angle ALM=\angle LAM+\angle ALM=\angle LMC=30^\circ$. Best regards, sunken rock

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