In a `DeltaABC`, if `b =(sqrt3-1) a and angle C=30^(@),` then the value of (A-B) is equal to (All symbols used have usual meaning in the triangel.)

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the given information
We are given:
- \( b = (\sqrt{3} - 1)a \)
- \( \angle C = 30^\circ \)

### Step 2: Use the angle sum property of a triangle
In triangle \( ABC \), the sum of the angles is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting the known value of \( \angle C \):
\[ \angle A + \angle B + 30^\circ = 180^\circ \]
This simplifies to:
\[ \angle A + \angle B = 150^\circ \]

### Step 3: Apply the Law of Sines
According to the Law of Sines:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
From this, we can express \( \frac{b}{a} \):
\[ \frac{b}{a} = \frac{\sin B}{\sin A} \]
Substituting \( b = (\sqrt{3} - 1)a \):
\[ \frac{\sqrt{3} - 1}{1} = \frac{\sin B}{\sin A} \]

### Step 4: Rationalize the expression
We have:
\[ \frac{\sin B}{\sin A} = \sqrt{3} - 1 \]
To rationalize, we multiply by the conjugate:
\[ \frac{\sin B}{\sin A} = \frac{(\sqrt{3} - 1)(\sqrt{3} + 1)}{\sqrt{3} + 1} = \frac{2}{\sqrt{3} + 1} \]

### Step 5: Simplify further
Now, we can express this as:
\[ \frac{\sin B}{\sin A} = \frac{2}{\sqrt{3} + 1} \]
Dividing both sides by \( 2 \):
\[ \frac{\sin B}{\sin A} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3} + 1}{2} \]

### Step 6: Identify angles using sine values
We know:
\[ \frac{\sin B}{\sin A} = \frac{\sin 45^\circ}{\sin 105^\circ} \]
This implies:
\[ B = 45^\circ \quad \text{and} \quad A = 105^\circ \]

### Step 7: Calculate \( A - B \)
Now, we can find \( A - B \):
\[ A - B = 105^\circ - 45^\circ = 60^\circ \]

### Final Answer
Thus, the value of \( A - B \) is:
\[ \boxed{60^\circ} \]

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