jmc

We draw segment $\overline{CD}$ as shown parallel to lines $l$ and $k$.



Since $\overline{CD}\parallel l$, we have $\angle ACD=180^{\circ }-\angle A=60^{\circ }$, so $\angle DCB=\angle ACB-\angle ACD=80^{\circ }-60^{\circ }=20^{\circ }$. Since $\overline{CD}\parallel k$, we have $\angle B=180^{\circ }-\angle DCB=\boxed{160^{\circ }}$.