解：如图，作点$M$关于$OA$的对称点$C，$点$M$关于$OB$的对称点$D，$连接$CD$交$OA$于点$E，$交$OB$于点$F，$连接$EM，$$FM，$$OC，$$OD.$
易得$CE = ME，$$DF = MF，$$OC = OM = OD，$$\angle COA=\angle MOA = 10^{\circ}，$$\angle DOB=\angle MOB = 30^{\circ}，$由两点之间线段最短可得，此时$\triangle MEF$的周长最小，为$ME + MF + EF = CE + DF + EF = CD.$
$\therefore\angle DOM = 60^{\circ}，$$\angle COM = 20^{\circ}，$$\angle COD=\angle DOM+\angle COM = 80^{\circ}.$
$\therefore\triangle DOM$为等边三角形.
$\therefore\angle OMD=\angle ODM = 60^{\circ}.$
$\because OC = OD = OM，$
$\therefore\angle ODC=\angle OCD=\frac{1}{2}(180^{\circ}-\angle COD)=50^{\circ}.$
$\therefore\angle FDM=\angle ODM-\angle ODC = 10^{\circ}.$
$\because DF = MF，$
$\therefore\angle FMD=\angle FDM = 10^{\circ}.$
$\therefore\angle OMF=\angle OMD-\angle FMD = 50^{\circ}.$
$\because OC = OM，$
$\therefore\angle OMC=\angle OCM=\frac{1}{2}(180^{\circ}-\angle COM)=80^{\circ}.$
$\therefore\angle MCE=\angle OCM-\angle OCD = 30^{\circ}.$
$\because CE = ME，$
$\therefore\angle CME=\angle MCE = 30^{\circ}.$
$\therefore\angle EMO=\angle CMO-\angle CME = 50^{\circ}.$
$\therefore\angle EMF=\angle EMO+\angle OMF = 100^{\circ}.$
