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Question: In a triangle \[ABC,\]\(b = 2,{\text{ }}B = {30^0}\), then the area of the circumcircle of the trian...

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

Explanation

Solution

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:
asinA=bsinB=csinC=2R\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R, where a,b,ca,b,c are the sides of triangle and RR is the radius of the circle.
Now, B=300B = {30^0} is given,
\therefore Taking bsinB=2R\dfrac{b}{{\sin B}} = 2R and putting the values of bb and sinB\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 =πR2 = \pi {R^2}
=π(2)2 =4π  = \pi {\left( 2 \right)^2} \\\ = 4\pi \\\
\therefore Correct option is (C).\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.