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!