Question

Two sides of a triangle are 2 and $\sqrt{3}$ and the included angle is ${30}^{\prime }$ then the in-radius r of the triangle is equal

• A. $\frac{1}{4}(\sqrt{3}-1)$
• B. $\frac{1}{2}(\sqrt{3}+1)$
• C. $\frac{1}{2}(\sqrt{3}-1)$
• D. $\frac{1}{4}(\sqrt{3}+1)$

Answer 1

The correct option is C

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

We know that $r=(s-a)tan\frac{A}{2}$.
Let the given sides be $b$ and $c$ and $A={30}^{\circ }$ then

$a^2=b^2+c^2-2bc\cos A\text{by cosine rule}=4+3-4\sqrt{3}\cos {30}^{\circ }{a}^2=7-4\sqrt{3}\times \frac{\sqrt{3}}{2}=7-6=1\therefore a^2=1\Rightarrow a=12s=a+b+c=1+2+\sqrt{3}=3+\sqrt{3}\therefore 2s-2a=3+\sqrt{3}-2\Rightarrow (s-a)=\frac{\sqrt{3}+1}{2}A={30}^{\circ }\Rightarrow \frac{A}{2}={15}^{\circ }\therefore tan\frac{A}{2}=tan{15}^{\circ }=\frac{\sqrt{3}-1}{\sqrt{3}+1}\therefore r=(s-a)tan\frac{A}{2}=\frac{\sqrt{3}+1}{2}.\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{1}{2}(\sqrt{3}-1)$