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By the Pythagorean Theorem, $$\begin{array}{rl}Y{Z}^2& =Y{X}^2+X{Z}^2\\ & =60^2+80^2\\ & =3600+6400\\ & =10000,\end{array}$$ so $YZ=100.$

(We could also have found $YZ$ without using the Pythagorean Theorem by noticing that $△XYZ$ is a right-angled triangle with its right-angle at $X$ and $XY=60=3\cdot 20$ and $XZ=80=4\cdot 20.$ This means that $△XYZ$ is similar to a 3-4-5 triangle, and so $YZ=5\cdot 20=100.$)

Since $△YXZ$ is right-angled at $X,$ its area is $$\frac{1}{2}\cdot 60\cdot 80=2400.$$ Since $XW$ is perpendicular to $YZ,$ then the area of $△YXZ$ is also equal to $$\frac{1}{2}\cdot 100\cdot XW=50XW.$$ Therefore, $50XW=2400,$ so $XW=48.$ By the Pythagorean Theorem, $$\begin{array}{rl}W{Z}^2& =80^2-48^2\\ & =6400-2304\\ & =4096.\end{array}$$ Thus, $WZ=\sqrt{4096}=\boxed{64}.$

An alternative solution comes by noticing that $△XZW$ and $△YZX$ are similar. Therefore $$\frac{WZ}{XZ}=\frac{XZ}{YZ}$$ or $$\frac{WZ}{80}=\frac{80}{100}=\frac{4}{5}.$$ This tells us that $$WZ=\frac{4}{5}\cdot 80=\boxed{64}.$$