If the angles of a triangle are `30^(@)` and `45^(@),` and the included sise is `(sqrt3+ 1)` cm, then prove that the area of the triangle is `1/2 (sqrt3+1).`

Answer

To solve the problem, we need to find the area of a triangle with angles of \(30^\circ\) and \(45^\circ\) and an included side of \((\sqrt{3} + 1)\) cm. We will use the formula for the area of a triangle based on two sides and the included angle.

### Step 1: Identify the angles and the included side
Given:
- Angle A = \(30^\circ\)
- Angle B = \(45^\circ\)
- Included side (between angles A and B) = \(c = \sqrt{3} + 1\) cm

### Step 2: Calculate the third angle
Using the property that the sum of angles in a triangle is \(180^\circ\):
\[ \text{Angle C} = 180^\circ - (30^\circ + 45^\circ) = 180^\circ - 75^\circ = 105^\circ \]

### Step 3: Use the area formula
The area \(A\) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times a \times b \times \sin(C) \]
Where \(a\) and \(b\) are the sides adjacent to angle \(C\), and \(C\) is the included angle.

### Step 4: Apply the Law of Sines to find sides \(a\) and \(b\)
Using the Law of Sines:
\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]
We can express \(a\) and \(b\) in terms of \(c\):
\[ a = c \cdot \frac{\sin(A)}{\sin(C)} \quad \text{and} \quad b = c \cdot \frac{\sin(B)}{\sin(C)} \]

Substituting the known values:
- \(A = 30^\circ\), \(B = 45^\circ\), \(C = 105^\circ\), and \(c = \sqrt{3} + 1\):
\[ a = (\sqrt{3} + 1) \cdot \frac{\sin(30^\circ)}{\sin(105^\circ)} \quad \text{and} \quad b = (\sqrt{3} + 1) \cdot \frac{\sin(45^\circ)}{\sin(105^\circ)} \]

### Step 5: Calculate the sine values
Using known sine values:
- \(\sin(30^\circ) = \frac{1}{2}\)
- \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\)
- \(\sin(105^\circ) = \sin(60^\circ + 45^\circ) = \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ) = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}\)

### Step 6: Substitute back to find \(a\) and \(b\)
Substituting the sine values into the equations for \(a\) and \(b\):
\[ a = (\sqrt{3} + 1) \cdot \frac{\frac{1}{2}}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = (\sqrt{3} + 1) \cdot \frac{\sqrt{2}}{\sqrt{3} + 1} = \sqrt{2} \]
\[ b = (\sqrt{3} + 1) \cdot \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = (\sqrt{3} + 1) \cdot \frac{2}{\sqrt{3} + 1} = 2 \]

### Step 7: Calculate the area
Now substituting \(a\), \(b\), and \(\sin(105^\circ)\) into the area formula:
\[ A = \frac{1}{2} \cdot \sqrt{2} \cdot 2 \cdot \sin(105^\circ) = \sqrt{2} \cdot \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2} \]

### Conclusion
Thus, we have shown that the area of the triangle is:
\[ A = \frac{1}{2}(\sqrt{3} + 1) \]