In `Delta ABC angle A = (pi)/(6) & b:c= 2: sqrt(3) "then " angle B ` is

Options

A. `pi/3`
B. `pi/2 `
C. `pi/4 `
D. Non of these

Answer

To solve the problem, we will use the Law of Cosines and the given information about triangle ABC.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Angle A = \( \frac{\pi}{6} \) (which is 30 degrees)
- The ratio of sides \( b:c = 2:\sqrt{3} \)

2. **Express the Sides in Terms of a Variable:**
- Let \( b = 2k \) and \( c = \sqrt{3}k \) for some positive value \( k \).

3. **Use the Law of Cosines:**
The Law of Cosines states:
\[ a^2 = b^2 + c^2 - 2bc \cos A \]
Substitute the known values:
\[ a^2 = (2k)^2 + (\sqrt{3}k)^2 - 2(2k)(\sqrt{3}k) \cos\left(\frac{\pi}{6}\right) \]

4. **Calculate \( \cos\left(\frac{\pi}{6}\right) \):**
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

5. **Substituting Values:**
\[ a^2 = 4k^2 + 3k^2 - 2(2k)(\sqrt{3}k) \left(\frac{\sqrt{3}}{2}\right) \]
Simplifying further:
\[ a^2 = 4k^2 + 3k^2 - 2(2k)(\sqrt{3}k) \cdot \frac{\sqrt{3}}{2} \]
\[ = 4k^2 + 3k^2 - 2k^2 \cdot 3 \]
\[ = 7k^2 - 6k^2 = k^2 \]

6. **Finding the Value of \( a \):**
\[ a^2 = k^2 \implies a = k \]

7. **Finding Angle B Using the Law of Cosines Again:**
Now we need to find angle B using:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substitute the values:
\[ \cos B = \frac{k^2 + (\sqrt{3}k)^2 - (2k)^2}{2(k)(\sqrt{3}k)} \]
\[ = \frac{k^2 + 3k^2 - 4k^2}{2k^2\sqrt{3}} = \frac{0}{2k^2\sqrt{3}} = 0 \]

8. **Finding Angle B:**
Since \( \cos B = 0 \):
\[ B = \frac{\pi}{2} \]

### Final Answer:
Angle B is \( \frac{\pi}{2} \) radians.