Question

In a $\Delta ABC,b=2,c=\sqrt{3},\angle A={30}^{\circ }$. Evaluate inradius of $\Delta ABC$.

Answer 1

Given $b=2,c=\sqrt{3},\angle A={30}^o$

Now, $\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos {30}^o=\frac{2^2+(\sqrt{3}{)}^2-a^2}{4\sqrt{3}}$

$\frac{\sqrt{3}}{2}=\frac{4+3-a^2}{4\sqrt{3}}$

$7-a^2=6$

$\Rightarrow a^2=1$

$\Rightarrow a=1$ (Since, length of side of triangle is positive. )

Now, Inradius $r=\frac{\Delta }{s}$

where $\Delta$ is area of triangle and $s$ is semi-perimeter of triangle.

So, $s=\frac{a+b+c}{2}$

$s=\frac{1+2+\sqrt{3}}{2}$

$s=\frac{3+\sqrt{3}}{2}$

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

$\Delta =\sqrt{\frac{(3+\sqrt{3})}{2}(\frac{3+\sqrt{3}}{2}-1)(\frac{3+\sqrt{3}}{2}-2)(\frac{3+\sqrt{3}}{2}-\sqrt{3})}$

$\Delta =\sqrt{\frac{(3+\sqrt{3})}{2}\left(\frac{1+\sqrt{3}}{2}\right)\left(\frac{-1+\sqrt{3}}{2}\right)\left(\frac{3-\sqrt{3}}{2}\right)}$

$\Delta =\sqrt{\frac{3^2-(\sqrt{3}{)}^2}{4}\left(\frac{(\sqrt{3}{)}^2-1^2}{4}\right)}$

$\Delta =\sqrt{\frac{9-3}{4}\left(\frac{3-1}{4}\right)}$

$\Delta =\sqrt{\frac{12}{16}}$

$\Delta =\frac{\sqrt{3}}{2}$

Substituting the values in the formula of inradius , we get

$r=\frac{\frac{\sqrt{3}}{2}}{\frac{(3+\sqrt{3})}{2}}$

$r=\frac{\sqrt{3}}{3+\sqrt{3}}$

$r=\frac{\sqrt{3}}{3+\sqrt{3}}\times \frac{3-\sqrt{3}}{3-\sqrt{3}}$

$r=\frac{\sqrt{3}(3-\sqrt{3})}{9-3}$

$r=\frac{3(\sqrt{3}-1)}{6}$

$r=\frac{\sqrt{3}-1}{2}$