Question

In a $△ABC$, 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 the value of circum radius of triangle $ABC$ is ?

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

$b=2c=\sqrt{3}$

$\Rightarrow \Rightarrow b=\sqrt{3}$

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

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

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

$\Rightarrow \frac{3\sqrt{3}}{2}=\frac{15}{4}-a^2$

$\Rightarrow a^2=\frac{15}{4}-\frac{\sqrt{3}}{4}\Rightarrow a=1.673278$

We know $\frac{a}{\sin a}=\frac{1}{2R}$

$\Rightarrow R=\frac{\sin a}{2a}=\frac{\frac{1}{2}}{2}=\frac{1}{4}$