In a `DeltaABC`, if `A=45^(@)`, `b=sqrt(6)`, a=2, then B=

Options

A. `30^(@)or150^(@)`
B. `60^(@)or120^(@)`
C. `45^(@)or135^(@)`
D. none of these

Answer

To solve the problem, we will use the Sine Rule in triangle ABC, where we know angle A, side a, and side b.

### Step-by-Step Solution:

1. **Write down the Sine Rule**:
The Sine Rule states that:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \]
Here, we will focus on the first two parts of the equation.

2. **Substitute the known values**:
We know:
- \( A = 45^\circ \)
- \( a = 2 \)
- \( b = \sqrt{6} \)

Thus, we can write:
\[ \frac{\sin 45^\circ}{2} = \frac{\sin B}{\sqrt{6}} \]

3. **Calculate \(\sin 45^\circ\)**:
We know that:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]

Substituting this value into the equation gives:
\[ \frac{\frac{1}{\sqrt{2}}}{2} = \frac{\sin B}{\sqrt{6}} \]

4. **Simplify the left side**:
The left side simplifies to:
\[ \frac{1}{2\sqrt{2}} = \frac{\sin B}{\sqrt{6}} \]

5. **Cross-multiply**:
Cross-multiplying gives:
\[ \sin B = \frac{\sqrt{6}}{2\sqrt{2}} \]

6. **Simplify \(\frac{\sqrt{6}}{2\sqrt{2}}\)**:
We can simplify this further:
\[ \sin B = \frac{\sqrt{3} \cdot \sqrt{2}}{2\sqrt{2}} = \frac{\sqrt{3}}{2} \]

7. **Find angle B**:
We know that:
\[ \sin B = \frac{\sqrt{3}}{2} \]
This corresponds to:
\[ B = 60^\circ \quad \text{or} \quad B = 120^\circ \]

### Final Answer:
Thus, the possible values for angle B are \( 60^\circ \) or \( 120^\circ \).

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