In a `Delta ABC, a =2b and |A-B| =(pi)/(3).` Determine the `angleC.`

Answer

To solve the problem, we need to determine the angle \( C \) in triangle \( ABC \) given that \( a = 2b \) and \( |A - B| = \frac{\pi}{3} \).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We know that:
- \( a = 2b \)
- \( |A - B| = \frac{\pi}{3} \)

2. **Using the Formula for the Difference of Angles:**
We can use the formula:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cdot \cot\left(\frac{C}{2}\right) \]
Since \( |A - B| = \frac{\pi}{3} \), we can write:
\[ A - B = \frac{\pi}{3} \quad \text{or} \quad B - A = \frac{\pi}{3} \]

3. **Substituting Values:**
Let's assume \( A - B = \frac{\pi}{3} \). Then:
\[ \tan\left(\frac{\frac{\pi}{3}}{2}\right) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Now substituting \( a \) and \( b \):
\[ \frac{2b - b}{2b + b} \cdot \cot\left(\frac{C}{2}\right) = \frac{b}{3b} \cdot \cot\left(\frac{C}{2}\right) = \frac{1}{3} \cdot \cot\left(\frac{C}{2}\right) \]

4. **Setting Up the Equation:**
We have:
\[ \frac{1}{\sqrt{3}} = \frac{1}{3} \cdot \cot\left(\frac{C}{2}\right) \]
Multiplying both sides by 3:
\[ \cot\left(\frac{C}{2}\right) = 3 \cdot \frac{1}{\sqrt{3}} = \sqrt{3} \]

5. **Finding \( C \):**
We know that:
\[ \cot\left(\frac{C}{2}\right) = \sqrt{3} \]
This implies:
\[ \frac{C}{2} = \frac{\pi}{6} \quad \text{(since } \cot\left(\frac{\pi}{6}\right) = \sqrt{3}\text{)} \]
Therefore:
\[ C = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \]

### Final Answer:
Thus, the value of angle \( C \) is:
\[ C = \frac{\pi}{3} \]