Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

Given $b=2,c=\sqrt{3},\angle A={30}^0$

then the inradius of $\Delta ABC$ is?

• A. $\frac{\sqrt{3}-1}{2}$
• B. $\frac{\sqrt{3}+1}{2}$
• C. $\frac{\sqrt{3}-1}{4}$
• D. none of these

Answer 1

Answer: (A)

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

$\Delta =\frac{1}{2}\times 2\times \sqrt{3}\sin {30}^{\circ }=\frac{\sqrt{3}}{2}$

$\sin \frac{A}{2}=\sqrt{\frac{(s-2)(s-\sqrt{3})}{2\sqrt{3}}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\Rightarrow (s-2)(s-\sqrt{3})=\frac{(4-2\sqrt{3})\left(2\sqrt{3}\right)}{8}$

$\Rightarrow s^2-(2+\sqrt{3})s+2\sqrt{3}=\sqrt{3}-\frac{3}{2}$

$\Rightarrow s^2-(2+\sqrt{3})s+\sqrt{3}+\frac{3}{2}=0$

$s=\frac{(2+\sqrt{3})\pm \sqrt{4+3+4\sqrt{3}-4\times (\sqrt{3}+\frac{3}{2})}}{2}$

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

$[s\ne \frac{1+\sqrt{3}}{2}\because s=\frac{2+\sqrt{3}+a}{2}\text{would mean}a=-1]$

Now, $r=\frac{\Delta }{s}=\frac{\frac{\sqrt{3}}{2}}{\frac{3+\sqrt{3}}{2}}=\frac{\sqrt{3}-1}{2}$