Question

In a \(\triangle ABC\), if \(b = 2\), \(\angle B = 30^\circ\), then the area of the circumcircle of \(\triangle ABC\) (in sq units) is

• A. \(\pi\)
• B. \(2\pi\)
• C. \(4\pi\)
• D. \(6\pi\)

Hint:

Sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\), where \(R\) is circumradius.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Use sine rule: \(\frac{b}{\sin B} = 2R\).

Step 2: Detailed Explanation:
Given \(b = 2\), \(\sin B = \sin 30^\circ = \frac{1}{2}\).
\(2R = \frac{b}{\sin B} = \frac{2}{1/2} = 4 \implies R = 2\).
Area of circumcircle = \(\pi R^2 = \pi \times 4 = 4\pi\).

Step 3: Final Answer:
Option (C) \(4\pi\).