#### Answer

$-2$

#### Work Step by Step

First, we use the division theorem to divide the absolute values and subtract the arguments:
$\frac{10(\cos50^{\circ}+i\sin50^{\circ})}{5(\cos230^{\circ}+i\sin 230^{\circ})}
\\=2(\cos (50^{\circ}-230^{\circ})+i\sin(50^{\circ}-230^{\circ}))
\\=2(\cos-180^{\circ}+i\sin-180^{\circ})$
Since we know that $\cos-180^{\circ}=\cos180^{\circ}=-1$ and $\sin-180^{\circ}=-\sin180^{\circ}=0$, we can substitute these values in the expression and simplify:
$2(\cos-180^{\circ}+i\sin-180^{\circ})
\\=2(-1)
\\=-2$