In `DeltaABC`,
If the two angles of a triangle are `30^(@) and 45^(@)` and the included side is `(sqrt(3)+1)` cm, find the area of the triangle.

Answer

To find the area of triangle ABC where the angles are \(30^\circ\) and \(45^\circ\) and the included side \(C\) is \((\sqrt{3} + 1)\) cm, we can follow these steps:

### Step 1: Determine the Third Angle
Since the sum of angles in a triangle is \(180^\circ\), we can find angle \(C\):
\[ C = 180^\circ - A - B = 180^\circ - 30^\circ - 45^\circ = 105^\circ \]

### Step 2: Use the Sine Rule to Find Side \(B\)
We can use the sine rule, which states:
\[ \frac{A}{\sin A} = \frac{B}{\sin B} = \frac{C}{\sin C} \]
We want to find side \(B\), so we can set up the equation:
\[ \frac{B}{\sin B} = \frac{C}{\sin C} \]
Substituting the known values:
\[ B = \frac{\sin B}{\sin C} \cdot C \]
Where \(B = 45^\circ\) and \(C = \sqrt{3} + 1\).

### Step 3: Calculate \(\sin 45^\circ\) and \(\sin 105^\circ\)
Using known values:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \]
Calculating \(\sin 105^\circ\):
\[ \sin 105^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

### Step 4: Substitute to Find \(B\)
Now substituting back into the sine rule:
\[ B = \frac{\sin 45^\circ}{\sin 105^\circ} \cdot C = \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3} + 1}{2\sqrt{2}}} \cdot (\sqrt{3} + 1) \]
This simplifies to:
\[ B = \frac{2}{\sqrt{3} + 1} \cdot (\sqrt{3} + 1) = 2 \]

### Step 5: Calculate the Area of Triangle ABC
The area \(A\) of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \cdot B \cdot C \cdot \sin A \]
Substituting the known values:
\[ \text{Area} = \frac{1}{2} \cdot 2 \cdot (\sqrt{3} + 1) \cdot \sin 30^\circ \]
Since \(\sin 30^\circ = \frac{1}{2}\):
\[ \text{Area} = \frac{1}{2} \cdot 2 \cdot (\sqrt{3} + 1) \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \text{ cm}^2 \]

### Final Answer
Thus, the area of triangle ABC is:
\[ \text{Area} = \frac{\sqrt{3} + 1}{2} \text{ cm}^2 \]

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