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To solve the problem, let's start by analyzing the given information:

1. The angle \(\alpha\) is between \(0\) and \(\pi\).
2. \(\tan(\alpha) = -\frac{1}{\sqrt{3}}\).

### Step 1: Determine the Quadrant of \(\alpha\)
Given that \(\tan(\alpha)\) is negative and the angle \(\alpha\) is between \(0\) and \(\pi\), we know that \(\alpha\) must be in the second quadrant because:
- In the first quadrant (\(0 < \alpha < \frac{\pi}{2}\)), both sine and cosine are positive, so tangent would be positive.
- In the second quadrant (\(\frac{\pi}{2} < \alpha < \pi\)), sine is positive and cosine is negative, so tangent would be negative.

### Step 2: Use the Identity for Tangent
\[
\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}
\]
Given that \(\tan(\alpha) = -\frac{1}{\sqrt{3}}\), we know that:
\[
\frac{\sin(\alpha)}{\cos(\alpha)} = -\frac{1}{\sqrt{3}}
\]
This implies that:
\[
\sin(\alpha) = -\frac{1}{\sqrt{3}} \cdot \cos(\alpha)
\]

### Step 3: Use the Pythagorean Identity
We use the identity:
\[
\sin^2(\alpha) + \cos^2(\alpha) = 1
\]
Substitute \(\sin(\alpha) = -\frac{1}{\sqrt{3}} \cdot \cos(\alpha)\) into the equation:
\[
\left(-\frac{1}{\sqrt{3}} \cdot \cos(\alpha)\right)^2 + \cos^2(\alpha) = 1
\]
\[
\frac{1}{3} \cos^2(\alpha) + \cos^2(\alpha) = 1
\]
\[
\frac{4}{3} \cos^2(\alpha) = 1
\]
\[
\cos^2(\alpha) = \frac{3}{4}
\]
\[
\cos(\alpha) = \pm \frac{\sqrt{3}}{2}
\]
Since \(\alpha\) is in the second quadrant, \(\cos(\alpha)\) should be negative:
\[
\cos(\alpha) = -\frac{\sqrt{3}}{2}
\]

### Step 4: Find \(\sin(\alpha)\)
Now, substitute \(\cos(\alpha) = -\frac{\sqrt{3}}{2}\) back into \(\sin(\alpha) = -\frac{1}{\sqrt{3}} \cdot \cos(\alpha)\):
\[
\sin(\alpha) = -\frac{1}{\sqrt{3}} \cdot \left(-\frac{\sqrt{3}}{2}\right)
\]
\[
\sin(\alpha) = \frac{1}{2}
\]

### Answer
The correct answer is \(\boxed{a. \frac{1}{2}}\).

Would you like more details or have any other questions?

### Follow-up Questions:
1. What is the general relationship between sine, cosine, and tangent in different quadrants?
2. How would the solution change if the angle \(\alpha\) were in the third or fourth quadrant?
3. Can \(\tan(\alpha)\) ever equal \(-\frac{1}{\sqrt{3}}\) in the first quadrant?
4. What other trigonometric identities could be useful in solving such problems?
5. How do the signs of sine and cosine functions determine the sign of the tangent function?

**Tip:** When dealing with trigonometric identities, always consider the quadrant to determine the correct sign of the sine and cosine functions.

If for an angle α we have 0 < α < π and tan(α) = -1/√3, then sin(α) equals?

Explore the trigonometric calculation of sin(α) given that tan(α) = -1/√3 and 0 < α < π. This detailed step-by-step solution covers key trigonometric identities and quadrant rules, suitable for high school students in Grades 9-12.

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