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In a $\triangle A B C$ if $b=2, B=30^{\circ}$ then the area of the circumcircle of $\triangle A B C$ in square units is
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Verified Answer
The correct answer is:
$4 \pi$
We have,
$$
\frac{b}{\sin B}=2 R
$$
$\Rightarrow \quad R=\frac{B}{2 \sin B}=\frac{2}{2 \times \sin 30^{\circ}}=2$
$\therefore$ Area of circumcircle $=\pi R^{2}$
$=\pi(2)^{2}=4 \pi$
$$
\frac{b}{\sin B}=2 R
$$
$\Rightarrow \quad R=\frac{B}{2 \sin B}=\frac{2}{2 \times \sin 30^{\circ}}=2$
$\therefore$ Area of circumcircle $=\pi R^{2}$
$=\pi(2)^{2}=4 \pi$
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