The sides of a triangle are in the ratio `2:sqrt6:sqrt3+1`, then its angles are

Options

A. `45^@,45^@,90^@`
B. `60^@,30^@,90^@`
C. `45^@,60^@,75^@`
D. none

Answer

To find the angles of a triangle given the sides in the ratio \(2 : \sqrt{6} : \sqrt{3} + 1\), we can follow these steps:

### Step 1: Express the sides in a common form
Let the sides of the triangle be represented as:
- \( a = 2k \)
- \( b = \sqrt{6}k \)
- \( c = (\sqrt{3} + 1)k \)

where \( k \) is a positive constant.

### Step 2: Normalize the sides
To simplify the calculations, we can divide each side by \( 2k \):
- \( a' = \frac{a}{2k} = 1 \)
- \( b' = \frac{b}{2k} = \frac{\sqrt{6}}{2} \)
- \( c' = \frac{c}{2k} = \frac{\sqrt{3} + 1}{2} \)

Now, we have the normalized sides as:
- \( a' = 1 \)
- \( b' = \frac{\sqrt{6}}{2} \)
- \( c' = \frac{\sqrt{3} + 1}{2} \)

### Step 3: Use the sine rule
According to the sine rule, the ratio of the sides of a triangle is equal to the ratio of the sines of the opposite angles:
\[ \frac{a'}{\sin A} = \frac{b'}{\sin B} = \frac{c'}{\sin C} \]

### Step 4: Calculate the angles using the sine values
Now, we will find the angles corresponding to the sides:
1. For \( a' = 1 \):
\[ \sin A = 1 \implies A = 90^\circ \]

2. For \( b' = \frac{\sqrt{6}}{2} \):
\[ \sin B = \frac{\sqrt{6}}{2} \implies B = 60^\circ \]

3. For \( c' = \frac{\sqrt{3} + 1}{2} \):
To find angle \( C \):
\[ \sin C = \frac{\sqrt{3} + 1}{2} \]
We can find \( C \) using the sine inverse function:
\[ C = \sin^{-1}\left(\frac{\sqrt{3} + 1}{2}\right) \]

### Step 5: Calculate the angles
Since the sum of angles in a triangle is \( 180^\circ \):
\[ C = 180^\circ - A - B = 180^\circ - 90^\circ - 60^\circ = 30^\circ \]

### Conclusion
Thus, the angles of the triangle are:
- \( A = 90^\circ \)
- \( B = 60^\circ \)
- \( C = 30^\circ \)