Let's call the longest side of the triangle $x$. Since we know the measure of the smallest angle is opposite the smallest side, we can set up the following proportion: $$\frac{16}{\sin 35^\circ}=\frac{x}{\sin 80^\circ}.$$Now we can find $\sin 35$ and $\sin 80^\circ$. We have $\sin 35^\circ=\sin (45-10)^\circ = \sin 45^\circ \cos 10^\circ - \cos 45^\circ \sin 10^\circ = \frac{\sqrt{2}}{2}\left(\cos 10^\circ - \sin 10^\circ\right)$. We also have $\sin 80^\circ=\sin (45+35)^\circ=\sin 45^\circ \cos 35^\circ + \cos 45^\circ \sin 35^\circ = \frac{\sqrt{2}}{2}\left(\cos 35^\circ + \sin 35^\circ\right)$. Substituting these values into our proportion, we have \begin{align*}
\frac{16}{\frac{\sqrt{2}}{2}\left(\cos 10^\circ - \sin 10^\circ\right)}&=\frac{x}{\frac{\sqrt{2}}{2}\left(\cos 35^\circ + \sin 35^\circ\right)}\\
32&=\frac{x}{\frac{\sqrt{2}}{2}\left(\cos 35^\circ + \sin 35^\circ\right)}\\
x&=\boxed{64}.
\end{align*}