If the angles `alpha, beta, gamma` of a triangle satisfy the relation,
`sin((alpha-beta)/(2)) + sin ((alpha -gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)`, then
The measure of the smallest angle of the triangle is

Options

A. `30^(@)`
B. `40^(@)`
C. `45^(@)`
D. `50^(@)`

Explanation

To solve the problem, we need to analyze the given equation involving the angles of a triangle. Let's denote the angles of the triangle as \( \alpha, \beta, \) and \( \gamma \).

The equation given is:

\[ \sin\left(\frac{\alpha - \beta}{2}\right) + \sin\left(\frac{\alpha - \gamma}{2}\right) + \sin\left(\frac{3\alpha}{2}\right) = \frac{3}{2} \]

### Step 1: Use the property of angles in a triangle

Since \( \alpha, \beta, \gamma \) are the angles of a triangle, we know that:

\[ \alpha + \beta + \gamma = 180^\circ \]

### Step 2: Express \( \alpha \) in terms of \( \beta \) and \( \gamma \)

From the triangle angle sum property, we can express \( \alpha \) as:

\[ \alpha = 180^\circ - \beta - \gamma \]

### Step 3: Substitute \( \alpha \) into the equation

Now, substitute \( \alpha \) into the original equation:

\[ \sin\left(\frac{(180^\circ - \beta - \gamma) - \beta}{2}\right) + \sin\left(\frac{(180^\circ - \beta - \gamma) - \gamma}{2}\right) + \sin\left(\frac{3(180^\circ - \beta - \gamma)}{2}\right) = \frac{3}{2} \]

### Step 4: Simplify the sine terms

Using the sine subtraction identity, we can simplify:

1. \( \frac{(180^\circ - \beta - \gamma) - \beta}{2} = \frac{180^\circ - 2\beta - \gamma}{2} = 90^\circ - \frac{2\beta + \gamma}{2} \)
2. \( \frac{(180^\circ - \beta - \gamma) - \gamma}{2} = \frac{180^\circ - \beta - 2\gamma}{2} = 90^\circ - \frac{\beta + 2\gamma}{2} \)
3. \( \frac{3(180^\circ - \beta - \gamma)}{2} = 270^\circ - \frac{3(\beta + \gamma)}{2} \)

Thus, the equation becomes:

\[ \sin\left(90^\circ - \frac{2\beta + \gamma}{2}\right) + \sin\left(90^\circ - \frac{\beta + 2\gamma}{2}\right) + \sin\left(270^\circ - \frac{3(\beta + \gamma)}{2}\right) = \frac{3}{2} \]

Using the identity \( \sin(90^\circ - x) = \cos(x) \) and \( \sin(270^\circ - x) = -\cos(x) \):

\[ \cos\left(\frac{2\beta + \gamma}{2}\right) + \cos\left(\frac{\beta + 2\gamma}{2}\right) - \cos\left(\frac{3(\beta + \gamma)}{2}\right) = \frac{3}{2} \]

### Step 5: Analyze the equation

This equation is complex, but we can analyze the values of \( \beta \) and \( \gamma \) knowing they are angles of a triangle.

### Step 6: Assume \( \beta = \gamma \)

Since \( \beta \) and \( \gamma \) are equal in many triangles, let's assume \( \beta = \gamma \):

Then \( \alpha + 2\beta = 180^\circ \) implies \( \alpha = 180^\circ - 2\beta \).

Substituting back into the equation gives:

\[ \sin\left(\frac{(180^\circ - 2\beta) - \beta}{2}\right) + \sin\left(\frac{(180^\circ - 2\beta) - \beta}{2}\right) + \sin\left(\frac{3(180^\circ - 2\beta)}{2}\right) = \frac{3}{2} \]

This simplifies to:

\[ 2\sin\left(90^\circ - \frac{3\beta}{2}\right) + \sin(270^\circ - 3\beta) = \frac{3}{2} \]

### Step 7: Solve for \( \beta \)

By solving this equation, we find that \( \beta = 40^\circ \) and hence \( \gamma = 40^\circ \) and \( \alpha = 100^\circ \).

### Conclusion

The smallest angle in the triangle is:

\[ \boxed{40^\circ} \]