Find the angles of the triangle whose sides are `3+ sqrt3, 2sqrt3 and sqrt6.`

Answer

To find the angles of the triangle with sides \( a = 3 + \sqrt{3} \), \( b = 2\sqrt{3} \), and \( c = \sqrt{6} \), we will use the cosine rule. The cosine rule states:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 1: Calculate \( \cos C \)

Using the formula for \( \cos C \):

\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

First, we need to calculate \( a^2 \), \( b^2 \), and \( c^2 \):

- \( a^2 = (3 + \sqrt{3})^2 = 9 + 6\sqrt{3} + 3 = 12 + 6\sqrt{3} \)
- \( b^2 = (2\sqrt{3})^2 = 4 \times 3 = 12 \)
- \( c^2 = (\sqrt{6})^2 = 6 \)

Now substituting these values into the formula:

\[ \cos C = \frac{(12 + 6\sqrt{3}) + 12 - 6}{2(3 + \sqrt{3})(2\sqrt{3})} \]

Calculating the numerator:

\[ 12 + 6\sqrt{3} + 12 - 6 = 18 + 6\sqrt{3} \]

Calculating the denominator:

\[ 2(3 + \sqrt{3})(2\sqrt{3}) = 4\sqrt{3}(3 + \sqrt{3}) = 12\sqrt{3} + 4 \times 3 = 12\sqrt{3} + 12 \]

Thus,

\[ \cos C = \frac{18 + 6\sqrt{3}}{12\sqrt{3} + 12} \]

### Step 2: Simplifying \( \cos C \)

Factoring out 6 from the numerator and 12 from the denominator:

\[ \cos C = \frac{6(3 + \sqrt{3})}{12(1 + \sqrt{3})} = \frac{1}{2} \cdot \frac{3 + \sqrt{3}}{1 + \sqrt{3}} \]

Now, since \( \cos C = \frac{\sqrt{3}}{2} \), we find:

\[ C = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ \]

### Step 3: Calculate \( \cos B \)

Using the formula for \( \cos B \):

\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

Substituting the values:

\[ \cos B = \frac{(12 + 6\sqrt{3}) + 6 - 12}{2(3 + \sqrt{3})(\sqrt{6})} \]

Calculating the numerator:

\[ 12 + 6\sqrt{3} + 6 - 12 = 6 + 6\sqrt{3} \]

Calculating the denominator:

\[ 2(3 + \sqrt{3})(\sqrt{6}) = 2\sqrt{6}(3 + \sqrt{3}) = 6\sqrt{6} + 2\sqrt{18} = 6\sqrt{6} + 6\sqrt{2} \]

Thus,

\[ \cos B = \frac{6(1 + \sqrt{3})}{6(\sqrt{6} + \sqrt{2})} = \frac{1 + \sqrt{3}}{\sqrt{6} + \sqrt{2}} \]

### Step 4: Calculate \( A \)

Using the angle sum property of triangles:

\[ A + B + C = 180^\circ \]

Substituting the known values:

\[ A + 45^\circ + 30^\circ = 180^\circ \]

Thus,

\[ A = 180^\circ - 75^\circ = 105^\circ \]

### Conclusion

The angles of the triangle are:

- \( A = 105^\circ \)
- \( B = 45^\circ \)
- \( C = 30^\circ \)