Question

In a $\Delta ABC,$ let $a,b$ and $c$ denote the length of sides opposite to vertices $A,B$ and $C$ respectively. If $b=2,$ $c=\sqrt{3}$ and $\angle BAC=\frac{\pi }{6},$ then value of circumradius of triangle $ABC$ is

Answer 1

$\cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{4+3-a^2}{4\sqrt{3}}$
$\Rightarrow a=1$

Now, $R=\frac{a}{2\sin A}=\frac{1}{2\sin \left(\frac{\pi }{6}\right)}=1$