In a `DeltaABC`, if `angleA=45^(@). angleB=60^(@) and angleC=75^(@)`, find the ratio of its sides.

Answer

To find the ratio of the sides of triangle ABC given the angles, we can use the sine rule, which states that the ratio of the sides of a triangle is proportional to the sine of the angles opposite those sides.

### Step-by-Step Solution:

1. **Identify the Angles**:
- Given:
- \( \angle A = 45^\circ \)
- \( \angle B = 60^\circ \)
- \( \angle C = 75^\circ \)

2. **Use the Sine Rule**:
- According to the sine rule, the ratio of the sides opposite to the angles is given by:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
- Here, \( a \) is the side opposite angle A, \( b \) is the side opposite angle B, and \( c \) is the side opposite angle C.

3. **Calculate the Sine Values**:
- Calculate \( \sin A \), \( \sin B \), and \( \sin C \):
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 75^\circ = \sin(45^\circ + 30^\circ) \)
- Using the sine addition formula:
\[ \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \]
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \)
- Thus,
\[ \sin 75^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

4. **Set Up the Ratios**:
- Now we can express the ratios of the sides:
\[ \frac{a}{\sin 45^\circ} = \frac{b}{\sin 60^\circ} = \frac{c}{\sin 75^\circ} \]
- Substituting the sine values:
\[ \frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{\sqrt{3} + 1}{2\sqrt{2}}} \]

5. **Express the Ratios**:
- Let \( k \) be a constant such that:
\[ a = k \cdot \frac{1}{\sqrt{2}}, \quad b = k \cdot \frac{\sqrt{3}}{2}, \quad c = k \cdot \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

6. **Final Ratio**:
- The ratio of the sides \( a : b : c \) can be expressed as:
\[ a : b : c = \frac{1}{\sqrt{2}} : \frac{\sqrt{3}}{2} : \frac{\sqrt{3} + 1}{2\sqrt{2}} \]
- To simplify, multiply each term by \( 2\sqrt{2} \):
\[ a : b : c = 2 : \sqrt{6} : (\sqrt{3} + 1) \]

### Conclusion:
Thus, the ratio of the sides of triangle ABC is:
\[ \boxed{2 : \sqrt{6} : (\sqrt{3} + 1)} \]