The tangent of angle between the lines whose intercepts on the axes are $4,-5$ and $5,-4$  respectively

Equation of the line having intercepts a,b  is $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ , and its slope is  $\mathrm{m}=-\frac{\mathrm{b}}{\mathrm{a}}$
Hence the slope of the line which makes intercepts  $4,-5$ is ${\mathrm{m}}_1=-\frac{5}{4}$  , and 

the slope of the line which makes intercepts $5,-4$ is ${\mathrm{m}}_2=-\frac{4}{5}$
 If $\theta$ be the acute angle between two lines having slopes ${\mathrm{m}}_1$  and ${\mathrm{m}}_2$ then  $\mathrm{tan\theta }=\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$
Hence, 

$\begin{array}{c}\tan \theta =\left|\frac{-\frac{5}{4}+\frac{4}{5}}{1+1}\right|\\ =\frac{9}{40}\end{array}$

Therefore, $\tan \theta =\boxed{0.23}$