 Given: sin A = 2/3,     90° < A < 180°
     Find: tan A/2


Given that \( \sin A = \frac{2}{3} \) and \( 90^\circ < A < 180^\circ \), we need to find \( \tan \frac{A}{2} \).

### Step 1: Find \( \cos A \)
Since \( \sin A = \frac{2}{3} \), we can use the Pythagorean identity to find \( \cos A \):
\[
\sin^2 A + \cos^2 A = 1
\]
\[
\left( \frac{2}{3} \right)^2 + \cos^2 A = 1
\]
\[
\frac{4}{9} + \cos^2 A = 1
\]
\[
\cos^2 A = 1 - \frac{4}{9} = \frac{5}{9}
\]
\[
\cos A = -\frac{\sqrt{5}}{3}
\]
(Note: \( \cos A \) is negative because \( A \) is in the second quadrant where cosine is negative.)

### Step 2: Use the half-angle identity for \( \tan \frac{A}{2} \)
The half-angle identity for \( \tan \frac{A}{2} \) is:
\[
\tan \frac{A}{2} = \frac{1 - \cos A}{\sin A}
\]
Substitute \( \cos A = -\frac{\sqrt{5}}{3} \) and \( \sin A = \frac{2}{3} \) into the formula:
\[
\tan \frac{A}{2} = \frac{1 - \left( -\frac{\sqrt{5}}{3} \right)}{\frac{2}{3}} = \frac{1 + \frac{\sqrt{5}}{3}}{\frac{2}{3}}
\]
Simplify the expression:
\[
\tan \frac{A}{2} = \frac{\frac{3 + \sqrt{5}}{3}}{\frac{2}{3}} = \frac{3 + \sqrt{5}}{2}
\]

Thus, \( \tan \frac{A}{2} = \frac{3 + \sqrt{5}}{2} \).

Would you like further details on any step?

### Related questions:
1. How can you derive the half-angle identities for other trigonometric functions?
2. Why is the cosine negative in the second quadrant?
3. How would the result change if \( A \) were in a different quadrant?
4. Can we apply this approach to other trigonometric equations involving half-angles?
5. How would the solution change if the angle were given in radians instead of degrees?

**Tip:** Always remember to check the quadrant of the angle when determining the sign of trigonometric values.

Given: sin A = 2/3, 90° < A < 180°. Find: tan A/2

Learn how to find tan(A/2) when sin A = 2/3 and the angle A is in the second quadrant (90° < A < 180°). This solution uses the Pythagorean identity and the half-angle identity for tangent. Suitable for students in Grades 10-12.

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