Question

# In a triangle $ABC,$$b = 2,{\text{ }}B = {30^0}$, then the area of the circumcircle of the triangle $ABC$ in sq. units is$\left( A \right).{\text{ }}\pi {\text{ }} \\ \left( B \right).{\text{ }}2\pi {\text{ }} \\ \left( C \right).{\text{ }}4\pi {\text{ }} \\ \left( D \right).{\text{ }}6\pi \\$

Hint: Here, First we will find the value of radius of circumcircle by using sine rule, then apply formula for area of a circle to get the required answer.

$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R$, where $a,b,c$ are the sides of triangle and $R$ is the radius of the circle.
Now, $B = {30^0}$ is given,
$\therefore$ Taking $\dfrac{b}{{\sin B}} = 2R$ and putting the values of $b$ and $\sin B$, we get
$\Rightarrow \dfrac{2}{{\sin {{30}^0}}} = 2R \\ \Rightarrow \dfrac{2}{{\left( {\dfrac{1}{2}} \right)}} = 2R{\text{ }}\left\{ {\because \sin {{30}^0} = \dfrac{1}{2}} \right\} \\ \Rightarrow 4 = 2R \\ \Rightarrow R = 2 \\$
Now, the area of circle $= \pi {R^2}$
$= \pi {\left( 2 \right)^2} \\ = 4\pi \\$
$\therefore$ Correct option is $\left( C \right).$