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

To solve for \(\sec(\theta)\), \(\sin(\theta)\), and \(\tan(\theta)\) based on the given right triangle:

### Step 1: Find the hypotenuse
We can use the Pythagorean theorem to find the hypotenuse \(c\) of the right triangle.

\[
c = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}
\]

So, the hypotenuse is \(\sqrt{61}\).

### Step 2: Calculate the trigonometric values

1. **\(\sin(\theta)\)**: 
   \[
   \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{\sqrt{61}}
   \]

2. **\(\tan(\theta)\)**: 
   \[
   \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{5}
   \]

3. **\(\sec(\theta)\)**: 
   \[
   \sec(\theta) = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{61}}{5}
   \]

### Final Answers:
- \(\sin(\theta) = \frac{6}{\sqrt{61}}\)
- \(\tan(\theta) = \frac{6}{5}\)
- \(\sec(\theta) = \frac{\sqrt{61}}{5}\)

Let me know if you'd like more details or explanations!

Here are some related questions you might find useful:

1. How do you rationalize the denominator of \(\frac{6}{\sqrt{61}}\)?
2. What is \(\cos(\theta)\) for this triangle?
3. How would you calculate \(\cot(\theta)\) based on \(\tan(\theta)\)?
4. How does the Pythagorean theorem work in general for right triangles?
5. What is the relationship between \(\sec(\theta)\) and \(\cos(\theta)\)?

**Tip**: Always remember that the reciprocal identities like \(\sec(\theta) = \frac{1}{\cos(\theta)}\) can be used to simplify trigonometric calculations!

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

Learn how to find the exact values of sec(θ), sin(θ), and tan(θ) using a right triangle with legs of 5 and 6 units. This guide covers key trigonometric concepts, including the Pythagorean theorem and basic trigonometric functions, suitable for students in Grades 9-12.

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