In a `Delta ABC, AB=12cm, angle ABC=30^(@), angle ACB=45^(@)` find the area of `Delta ABC` .

Answer

To find the area of triangle ABC given the sides and angles, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Side \( AB = 12 \, \text{cm} \)
- Angle \( \angle ABC = 30^\circ \)
- Angle \( \angle ACB = 45^\circ \)

2. **Calculate Angle \( \angle BAC \)**:
- Using the triangle angle sum property, we can find angle \( \angle BAC \):
\[ \angle BAC = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 30^\circ - 45^\circ = 105^\circ \]

3. **Use the Formula for Area of Triangle**:
- The area \( A \) of triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times a \times b \times \sin(C) \]
where \( a \) and \( b \) are two sides and \( C \) is the included angle between them. Here, we can use sides \( AB \) and \( AC \) (which we will find) with angle \( \angle BAC \).

4. **Find Side \( AC \)**:
- To find side \( AC \), we can use the Law of Sines:
\[ \frac{AB}{\sin(\angle ACB)} = \frac{AC}{\sin(\angle ABC)} \]
Rearranging gives:
\[ AC = \frac{AB \cdot \sin(\angle ABC)}{\sin(\angle ACB)} = \frac{12 \cdot \sin(30^\circ)}{\sin(45^\circ)} \]
- We know \( \sin(30^\circ) = \frac{1}{2} \) and \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \):
\[ AC = \frac{12 \cdot \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{12 \cdot 1}{\sqrt{2}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2} \, \text{cm} \]

5. **Find Side \( BC \)**:
- Now we can find side \( BC \) using the Law of Sines again:
\[ \frac{BC}{\sin(\angle BAC)} = \frac{AB}{\sin(\angle ACB)} \]
Rearranging gives:
\[ BC = \frac{AB \cdot \sin(\angle BAC)}{\sin(\angle ACB)} = \frac{12 \cdot \sin(105^\circ)}{\sin(45^\circ)} \]
- We know \( \sin(105^\circ) = \sin(180^\circ - 75^\circ) = \sin(75^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \):
\[ BC = \frac{12 \cdot \frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{2}}{2}} = \frac{12(\sqrt{6} + \sqrt{2})}{4} \cdot \frac{2}{\sqrt{2}} = \frac{12(\sqrt{6} + \sqrt{2}) \cdot 2}{4\sqrt{2}} = \frac{6(\sqrt{6} + \sqrt{2})}{\sqrt{2}} = 3(\sqrt{12} + 2) = 6\sqrt{3} + 6 \]

6. **Calculate the Area**:
- Now we can calculate the area using the formula:
\[ A = \frac{1}{2} \times AB \times AC \times \sin(\angle BAC) = \frac{1}{2} \times 12 \times 6\sqrt{2} \times \sin(105^\circ) \]
- Substituting \( \sin(105^\circ) \):
\[ A = \frac{1}{2} \times 12 \times 6\sqrt{2} \times \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{72\sqrt{2}(\sqrt{6} + \sqrt{2})}{8} = 9\sqrt{2}(\sqrt{6} + \sqrt{2}) = 9(\sqrt{12} + 2) = 9(2\sqrt{3} + 2) = 18\sqrt{3} + 18 \]

### Final Answer:
The area of triangle ABC is \( 18\sqrt{3} + 18 \, \text{cm}^2 \).