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The problem you provided is:

**Given**: \(\tan A = \frac{4}{3}\), with \(180^\circ < A < 270^\circ\) (i.e., in the third quadrant).

**Find**: \(\sin \frac{A}{2}\).

### Solution:

We will use the following steps:

1. **Determine \(\sin A\) and \(\cos A\)**:
   - Since \(\tan A = \frac{\sin A}{\cos A}\), and \(\tan A = \frac{4}{3}\), we can use a right triangle to find \(\sin A\) and \(\cos A\).
   - In the third quadrant, both sine and cosine are negative.

   Let:
   \[
   \sin A = -\frac{4}{5}, \quad \cos A = -\frac{3}{5}
   \]
   (These values come from using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\)).

2. **Apply the half-angle formula for sine**:
   The half-angle identity for sine is:
   \[
   \sin \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{2}}
   \]
   Since \(180^\circ < A < 270^\circ\), \(\frac{A}{2}\) is in the second quadrant, where sine is positive. Thus, we use the positive square root.

3. **Substitute \(\cos A = -\frac{3}{5}\)** into the half-angle formula:
   \[
   \sin \frac{A}{2} = \sqrt{\frac{1 - (-\frac{3}{5})}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}}
   \]

4. **Simplify**:
   \[
   \sin \frac{A}{2} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}
   \]

Thus, \(\sin \frac{A}{2} = \frac{2\sqrt{5}}{5}\).

Would you like further details on any step? Here are some related questions:

1. How do you find \(\cos A\) and \(\sin A\) using the Pythagorean identity?
2. Why is the half-angle formula used here, and how does it apply to this problem?
3. Can you explain how to determine the correct sign for the half-angle formula?
4. What is the relationship between the quadrant and the sign of \(\sin \frac{A}{2}\)?
5. How would this solution change if \(A\) were in a different quadrant?

**Tip**: Always consider the quadrant to decide the correct sign of trigonometric values!

Given: tan A = 4/3, 180° < A < 270°. Find: sin(A/2)

Find the sine of half an angle when tan A = 4/3, and A lies in the third quadrant (180° < A < 270°). This problem involves using trigonometric identities, including the Pythagorean identity and the half-angle formula. Suitable for students in Grades 10-12.

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