In a triangle ABC, if `(sqrt3-1)a = 2b, A = 3B`, then `/_C` is

Explanation

To solve the problem step by step, we will use the given information and properties of triangles.

### Step 1: Use the given relationships
We are given two relationships:
1. \((\sqrt{3} - 1)a = 2b\)
2. \(A = 3B\)

### Step 2: Apply the Law of Sines
Using the Law of Sines, we can express the sides in terms of the angles:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
From this, we can write:
\[ \frac{a}{b} = \frac{\sin A}{\sin B} \]

### Step 3: Substitute the relationship into the equation
From the first relationship, we can express \(a\) in terms of \(b\):
\[ a = \frac{2b}{\sqrt{3} - 1} \]
Now substituting this into the Law of Sines gives:
\[ \frac{\frac{2b}{\sqrt{3} - 1}}{b} = \frac{\sin A}{\sin B} \]
This simplifies to:
\[ \frac{2}{\sqrt{3} - 1} = \frac{\sin A}{\sin B} \]

### Step 4: Substitute \(A = 3B\)
Since \(A = 3B\), we can substitute this into the sine ratio:
\[ \frac{\sin(3B)}{\sin B} = \frac{2}{\sqrt{3} - 1} \]

### Step 5: Use the sine triple angle formula
Using the sine triple angle formula:
\[ \sin(3B) = 3\sin B - 4\sin^3 B \]
Thus, we can write:
\[ \frac{3\sin B - 4\sin^3 B}{\sin B} = \frac{2}{\sqrt{3} - 1} \]
This simplifies to:
\[ 3 - 4\sin^2 B = \frac{2}{\sqrt{3} - 1} \]

### Step 6: Solve for \(\sin B\)
Cross-multiplying gives:
\[ (3 - 4\sin^2 B)(\sqrt{3} - 1) = 2 \]
Expanding this and rearranging will yield a quadratic equation in terms of \(\sin^2 B\).

### Step 7: Calculate angles A and B
Once we find \(\sin B\), we can find \(B\) and subsequently \(A\) using \(A = 3B\).

### Step 8: Find angle C
Using the property that the sum of angles in a triangle is \(180^\circ\):
\[ C = 180^\circ - A - B \]
Substituting the values of \(A\) and \(B\) will give us angle \(C\).

### Final Answer
After performing the calculations, we find that:
\[ C = 120^\circ \]