The image contains a question that asks for the values of the six trigonometric functions based on a right triangle with the following side lengths:

• Opposite side to angle $\theta$: 7
• Adjacent side to angle $\theta$: 14
• Hypotenuse: It is not provided, but we can calculate it using the Pythagorean theorem.

To proceed:

1. Calculate the hypotenuse: $\text{Hypotenuse} = \sqrt{7^2 + 14^2} = \sqrt{49 + 196} = \sqrt{245} = 7\sqrt{5}$

Now we can calculate the six trigonometric functions.

1. Sine ($\sin \theta$):

$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{7}{7\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

2. Cosine ($\cos \theta$):

$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{14}{7\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

3. Tangent ($\tan \theta$):

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{7}{14} = \frac{1}{2}$

4. Cosecant ($\csc \theta$) (reciprocal of sine):

$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{5}}{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$

5. Secant ($\sec \theta$) (reciprocal of cosine):

$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{2\sqrt{5}}{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}$

6. Cotangent ($\cot \theta$) (reciprocal of tangent):

$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{1}{2}} = 2$

Summary of Answers:

• $\sin \theta = \frac{\sqrt{5}}{5}$
• $\cos \theta = \frac{2\sqrt{5}}{5}$
• $\tan \theta = \frac{1}{2}$
• $\csc \theta = \sqrt{5}$
• $\sec \theta = \frac{\sqrt{5}}{2}$
• $\cot \theta = 2$

Would you like further explanations on any of these calculations?

Here are some related questions to consider:

1. How does the Pythagorean theorem help in solving right triangle problems?
2. What are the reciprocal trigonometric functions and how are they related to the primary ones?
3. How can we use the values of sine and cosine to find angles?
4. What is the importance of rationalizing the denominator in trigonometric expressions?
5. How do trigonometric ratios apply to non-right triangles?

Tip: Always ensure you simplify trigonometric expressions, especially when radicals are involved.