Answer
$1.5(\cos 19^{\circ}+i\sin19^{\circ})$
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
$\frac{18(\cos 51^{\circ}+i\sin 51^{\circ})}{12(\cos 32^{\circ}+i\sin32^{\circ})}=\frac{18}{12}[\cos (51^{\circ}-32^{\circ})+i\sin(51^{\circ}-32^{\circ})]$
$=1.5(\cos 19^{\circ}+i\sin19^{\circ})$
