jmc

Let $n$ be the number of sides in the polygon. The sum of the interior angles in any $n$-sided polygon is $180(n-2)$ degrees. Since each angle in the given polygon measures $140^{\circ }$, the sum of the interior angles of this polygon is also $140n$. Therefore, we must have $$180(n-2)=140n.$$ Expanding the left side gives $180n-360=140n$, so $40n=360$ and $n=\boxed{9}$.

We might also have noted that each exterior angle of the given polygon measures $180^{\circ }-140^{\circ }=40^{\circ }$. The exterior angles of a polygon sum to $360^{\circ }$, so there must be $\frac{360^{\circ }}{40^{\circ }}=9$ of them in the polygon.