smc

Let the triangle be $△ABC$ with $\measuredangle BAC=60^{\circ }$ and $AC=80$, as in the diagram. Let $AB=x$. from the problem, we know that $BC=90-x$. Now, we can apply the Law of Cosines on $△ABC$ to solve for $x$: $$\begin{array}{rl}(90-x{)}^2& =6400+x^2-160\cos (60^{\circ })\\ x^2-180x+8100& =6400+x^2-80x\\ -100x& =-1700\\ x& =17\end{array}$$ Because the problem asks for the shortest side of the triangle and $90-x=73>17$, our answer is $\boxed{17}$.