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 \\
$

Answer 1

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.

Complete step-by-step answer:
Solving for the area of the circumcircle, we first need the radius of the circle.
In order to find the radius of the circle, we are using sine rule which states:
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R$, 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 $\left( C \right).$

Note: The Law of Sines (sine rule) is an important rule relating the sides and angles of any triangle. In order to find the area of circumcircle, we will first calculate the radius using sine rule, then apply formula for area of a circle.