The angle of a right angled triangle are $A.P.$ The ratio of the in-radius and the perimeter is

Let ABC be the right angled triangle whose angles

are in $A.P.$ Then,

Now, $A+B+C={180}^{\circ }\Rightarrow 3B={180}^{\circ }\Rightarrow B={60}^{\circ }$

So, let the angles be $A={30}^{\circ },B={60}^{\circ }$ and $C={90}^{\circ }$

$\begin{array}{l}\therefore \frac{a}{\sin {30}^{\circ }}=\frac{b}{\sin {60}^{\circ }}=\frac{c}{\sin {90}^{\circ }}=2R\\ \Rightarrow a=R,b=\sqrt{3}R\text{ and }c=2R\end{array}$

Also,

$\begin{array}{l}\mathrm{\Delta }=\frac{1}{2}ab\sin {90}^{\circ }=\frac{1}{2}ab=\frac{\sqrt{3}}{2}{R}^2\\ \therefore \frac{r}{2s}=\frac{\mathrm{\Delta }}{2s^2}\\ \Rightarrow \frac{r}{2s}=\frac{\sqrt{3}{R}^2}{4{\left({\displaystyle \frac{R+R\sqrt{3}+2R}{2}}\right)}^2}=\frac{\sqrt{3}}{{\left(1+2+\sqrt{3}\right)}^2}\\ \Rightarrow \frac{r}{2s}=\frac{\sqrt{3}}{{\left(\sqrt{3}\left(1+\sqrt{3}\right)\right)}^2}\\ \Rightarrow \frac{r}{2s}=\frac{\sqrt{3}}{3\left(4+2\sqrt{3}\right)}=\frac{1}{2\sqrt{3}\left(2+\sqrt{3}\right)}\\ \Rightarrow \frac{r}{2s}=\frac{2-\sqrt{3}}{2\sqrt{3}}\end{array}$