Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.
Given $b=2,c=\sqrt{3},\angle A={30}^{\circ },$ then inradius of $△ABC$ 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

The correct option is A $\frac{\sqrt{3}-1}{2}$

given $b=2,c=\sqrt{3},\angle A={30}^{\circ }$
From cosine rule
$\cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{4+3-a^2}{2\times 2\times \sqrt{3}}$
$\Rightarrow a=1$
$r=\frac{△}{s}=\frac{2bc\sin A}{2(a+b+c)}=\frac{2\times \sqrt{3}\times \sin {30}^{\circ }}{2(1+2+\sqrt{3})}=\frac{\sqrt{3}}{\sqrt{3}+3}=\frac{\sqrt{3}-1}{2}$
Ans: A