If the sides a triangle are in the ratio `2:sqrt6:(sqrt3 + 1)`, then the largest angle of the triangle will be

Options

A. `60^@`
B. `75^@`
C. `90^@`
D. `120^@`

Answer

To solve the problem, we need to find the largest angle of a triangle given the sides in the ratio \(2 : \sqrt{6} : (\sqrt{3} + 1)\). We can use the property that the ratio of the sides of a triangle is equal to the ratio of the sines of the angles opposite those sides.

### Step-by-step Solution:

1. **Identify the sides and their ratios**:
Let the sides of the triangle be \(a = 2k\), \(b = \sqrt{6}k\), and \(c = (\sqrt{3} + 1)k\) for some positive constant \(k\). The ratios of the sides are \(2 : \sqrt{6} : (\sqrt{3} + 1)\).

2. **Set up the sine ratio**:
According to the property of triangles, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
This implies:
\[ \frac{2k}{\sin A} = \frac{\sqrt{6}k}{\sin B} = \frac{(\sqrt{3} + 1)k}{\sin C} \]

3. **Simplify the ratios**:
We can ignore \(k\) since it is a common factor:
\[ \frac{2}{\sin A} = \frac{\sqrt{6}}{\sin B} = \frac{\sqrt{3} + 1}{\sin C} \]

4. **Express the sines in terms of the sides**:
From the ratios, we can express the sines:
\[ \sin A = \frac{2 \sin B}{\sqrt{6}} \quad \text{and} \quad \sin C = \frac{(\sqrt{3} + 1) \sin B}{\sqrt{6}} \]

5. **Use known sine values**:
We can assign angles based on known sine values:
- \(\sin 45^\circ = \frac{1}{\sqrt{2}}\)
- \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
- \(\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}\)

6. **Assign angles based on the sine values**:
- From the ratios, we can conclude:
- \(A\) corresponds to \(45^\circ\)
- \(B\) corresponds to \(60^\circ\)
- \(C\) corresponds to \(75^\circ\)

7. **Identify the largest angle**:
The largest angle among \(A\), \(B\), and \(C\) is \(C\), which is \(75^\circ\).

### Conclusion:
The largest angle of the triangle is \(75^\circ\).