The angles of a right angled triangle are in AP. The ratio of the inradius and the perimeter is

$\alpha -\beta ,\alpha ,\alpha +\beta$ be the angles since they are in A. P 

Let $\alpha -\beta +\alpha +\alpha +\beta ={180}^{\circ }\Rightarrow \alpha ={60}^{\circ }$

$\Rightarrow \text{ Angles are : }{30}^{\circ },{60}^{\circ },{90}^{\circ }(\mathrm{C},\mathrm{B},\text{ A respectively })$

$r=(b-a)\tan \frac{A}{2}$, where $A=\alpha +\beta \text{ (say) }$

$\begin{array}{l}r=s-a\\ \frac{r}{2s}=\frac{1}{2}-\frac{a}{2s}\end{array}$

$\begin{array}{l}=\frac{1}{2}-\frac{a}{a+b+c}\\ =\frac{1}{2}-\frac{\sin A}{\sum \sin A}\\ =\frac{2-\sqrt{3}}{2\sqrt{3}}\end{array}$