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

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

1. A

$\left(2+\sqrt{3}\right):2\sqrt{3}$

2. B

$\left(2+\sqrt{3}\right):\sqrt{3}$

3. C

$\left(2-\sqrt{3}\right):2\sqrt{3}$

4. D

$\left(2-\sqrt{3}\right):4\sqrt{3}$

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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

Let  Then,

Also,

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

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

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

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