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The given image shows a right triangle with a base of 7 units, a height of 8 units, and an unknown hypotenuse. We are tasked with finding the trigonometric values for \(\tan \theta\), \(\sec \theta\), and \(\sin \theta\), where \(\theta\) is the angle opposite the side of length 8.

### Step 1: Find the Hypotenuse
We can use the Pythagorean theorem to find the hypotenuse \(c\), where:
\[
c = \sqrt{7^2 + 8^2} = \sqrt{49 + 64} = \sqrt{113}
\]

So the hypotenuse \(c = \sqrt{113}\).

### Step 2: Calculate the Trigonometric Values
1. **\(\tan \theta\)** (opposite over adjacent):
\[
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{7}
\]

2. **\(\sec \theta\)** (hypotenuse over adjacent):
\[
\sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{113}}{7}
\]

3. **\(\sin \theta\)** (opposite over hypotenuse):
\[
\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{\sqrt{113}}
\]

### Final Answer:
- \(\tan \theta = \frac{8}{7}\)
- \(\sec \theta = \frac{\sqrt{113}}{7}\)
- \(\sin \theta = \frac{8}{\sqrt{113}}\)

Would you like more details on any step? Also, here are some follow-up questions for deeper understanding:

1. How can we rationalize the denominator of \(\sin \theta\)?
2. Can you derive \(\cos \theta\) from these values?
3. What are the other trigonometric functions (\(\cot \theta\), \(\csc \theta\)) for this triangle?
4. How would you calculate \(\theta\) using an inverse trigonometric function?
5. Why does the Pythagorean theorem apply to this scenario?

**Tip**: When working with trigonometric functions, always keep in mind which side corresponds to opposite, adjacent, and hypotenuse relative to the angle in question.

Find tan θ, sec θ, and sin θ, where θ is the angle shown in the figure. Give exact values, not decimal approximations.

Learn how to find the exact values of tan θ, sec θ, and sin θ for a right triangle. This problem involves calculating the hypotenuse using the Pythagorean theorem and applying trigonometric ratios. Suitable for high school students in Grades 9-12.

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