Answer
$3(\cos 50^{\circ}+i\sin50^{\circ})$
Work Step by Step
Recall: If $z_{1}=r_{1}(\cos\theta_{1}+i\sin\theta_{1})$ and $z_{2}=r_{2}(\cos\theta_{2}+i\sin\theta_{2})$, then
$\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})]$
Dividing according to the above formula, we get
$\frac{30(\cos 80^{\circ}+i\sin 80^{\circ})}{10(\cos 30^{\circ}+i\sin30^{\circ})}=\frac{30}{10}[\cos (80^{\circ}-30^{\circ})+i\sin(80^{\circ}-30^{\circ})]$
$=3(\cos 50^{\circ}+i\sin50^{\circ})$