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To calculate \(\tan 2\theta\) for the given triangle, let's proceed with the following steps:

### Step 1: Analyze the triangle
The triangle \(ABC\) is a right triangle where:
- \(AB = 8\) (opposite to \(\theta\))
- \(BC = 6\) (adjacent to \(\theta\))
- \(AC\) is the hypotenuse.

### Step 2: Calculate the hypotenuse \(AC\)
Using the Pythagorean theorem:
\[
AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10
\]

### Step 3: Determine \(\tan \theta\)
\(\tan \theta\) is given by:
\[
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{8}{6} = \frac{4}{3}
\]

### Step 4: Use the double angle identity for tangent
The double angle identity for tangent is:
\[
\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}
\]
Substitute \(\tan \theta = \frac{4}{3}\) into the identity:
\[
\tan 2\theta = \frac{2 \times \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}}
\]

### Step 5: Simplify the expression
\[
\tan 2\theta = \frac{\frac{8}{3}}{\frac{9 - 16}{9}} = \frac{\frac{8}{3}}{-\frac{7}{9}} = \frac{8}{3} \times -\frac{9}{7} = -\frac{72}{21} = -\frac{24}{7}
\]

Thus, \(\tan 2\theta = -\frac{24}{7}\).

Would you like further details on any of the steps or have any additional questions?

### Related Questions:
1. How do you calculate the sine of \(\theta\) in this triangle?
2. What is the cosine of \(\theta\) in this triangle?
3. How do you derive the formula for \(\tan 2\theta\)?
4. Can you calculate the angle \(\theta\) using the arctan function?
5. What are the trigonometric identities involving double angles?
6. How do you calculate the area of triangle \(ABC\)?
7. What is the length of the median from \(A\) to \(BC\)?
8. How would \(\tan 2\theta\) change if the triangle's dimensions were doubled?

### Tip:
When working with trigonometric identities, it's helpful to first calculate the basic trigonometric functions (like \(\tan \theta\)) before applying more complex identities like the double angle formulas.

Learn how to calculate tan 2θ in a right triangle using the double angle identity for tangent. Follow a detailed step-by-step solution including the application of the Pythagorean theorem. Suitable for Grades 10-12 students and anyone studying trigonometric identities.

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