If the angles of a right angled triangle are in AP., then the ratio of the in-radius and the perimeter, is

Solution:

Let ABC be a right angled triangle right angle at $B.$

Let the other angles be $A={90}^{\circ }-d$ and $C={90}^{\circ }-2d$. Then

$\begin{array}{l}A+B+C={180}^{\circ }\Rightarrow {270}^{\circ }-3d={180}^{\circ }\Rightarrow d={30}^{\circ }\\ A={60}^{\circ }\text{ and }C={30}^{\circ }\end{array}$

Let $AC =b.$ Then,

$\begin{array}{l}c=AB=AC\cos {60}^{\circ }=\frac{b}{2}\text{ and }a=BC=b\cos {30}^{\circ }=\frac{\sqrt{3}}{2}\\ \therefore 2s=a+b+c=\frac{\sqrt{3}b}{2}+b+\frac{b}{2}=\frac{(3+\sqrt{3})b}{2}\end{array}$

Also,

$\mathrm{\Delta }=\frac{1}{2}(BC\times AB)=\frac{1}{2}\left(\frac{\sqrt{3}b}{2}\times \frac{b}{2}\right)=\frac{\sqrt{3}{b}^2}{8}$

$\therefore$ Require ratio $=\frac{r}{2s}=\frac{\mathrm{\Delta }}{2s^2}$

$\begin{array}{l}=\frac{\frac{\sqrt{3}{b}^2}{8}}{2\left\{\frac{(3+\sqrt{3}{)}^2}{4}{b}^2\right\}}=\frac{\sqrt{3}}{(3+\sqrt{3}{)}^2}=\frac{\sqrt{3}}{3(\sqrt{3}+1{)}^2}\\ =\frac{(\sqrt{3}-1{)}^2}{4\sqrt{3}}=\frac{4-2\sqrt{3}}{4\sqrt{3}}=\frac{2-\sqrt{3}}{2\sqrt{3}}\end{array}$