解：由题意得$BC=\sqrt {AB^2-AC^2}=2\sqrt 2m$
$B'C=\sqrt {A'B'^2-A'C^2}=\sqrt 5m$
$\mathrm {tan}α=\frac {AC}{BC}=\frac {\sqrt 2}4$
$\mathrm {tan}α'=\frac {A'C}{B'C}=\frac {2\sqrt 5}5$
$\mathrm {tan}α$的值可以大于$100，$锐角$α$的正切值$\mathrm {tan}α$的范围为$\mathrm {tan}α>0$
