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

$\text{ Equation of the line having intercepts }a,b\text{ is }\frac{x}{a}+\frac{y}{b}=1,\text{ and its slope is }m=-\frac{b}{a}$

$\text{ Hence the slope of the line which makes intercepts }4,-5\text{ is }{m}_1=-\frac{(-5)}{4}=\frac{5}{4},\text{ and the slope of the }$

$\text{ line which makes intercepts }5,-4\text{ is }{m}_2=-\frac{(-4)}{5}=\frac{4}{5}$

$\text{ If }\theta \text{ be the acute angle between two lines having slopes }{m}_1\text{ and }{m}_2\text{ then }\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Hence, 

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

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