https://file.aiquickdraw.com/mathbot/file/1727066080551na97uxel.jpg

The problem in the image is asking for trigonometric ratios based on the triangle provided.

The triangle shows:
- Angle \( P \)
- \( RQ = 3 \) (adjacent to \( \angle P \))
- \( PQ = 4 \) (opposite to \( \angle P \))
- The hypotenuse \( PR \), calculated as \( q \).

To solve for \( q \) (the hypotenuse), the Pythagorean theorem is applied:

\[
q = \sqrt{p^2 + r^2}
\]
Where \( p = 3 \) and \( r = 4 \). Substituting in these values:

\[
q = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

So, the hypotenuse \( q = 5 \).

### Trigonometric Ratios for Angle \( P \):

1. **Sine** (\( \sin P \)): 
   \[
   \sin P = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{4}{5}
   \]

2. **Cosine** (\( \cos P \)):
   \[
   \cos P = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{RQ}{PR} = \frac{3}{5}
   \]

3. **Tangent** (\( \tan P \)):
   \[
   \tan P = \frac{\text{opposite}}{\text{adjacent}} = \frac{PQ}{RQ} = \frac{4}{3}
   \]

Let me know if you would like further clarification, more details on any step, or additional related topics explained!

### Here are 5 follow-up questions based on this:
1. How do you find the trigonometric ratios for angle \( R \) in this triangle?
2. Can you explain why the Pythagorean theorem is valid for right-angled triangles?
3. How would the trigonometric values change if \( PQ = 5 \) and \( RQ = 12 \)?
4. What is the significance of the sine and cosine values being fractions?
5. How can you use trigonometric ratios to find the missing angle \( P \)?

### Tip:
Always ensure you remember the basic trigonometric identities: sine, cosine, and tangent are defined with respect to the opposite, adjacent, and hypotenuse sides of a right triangle.

Tentukan perbandingan trigonometri di sudut R & P.

Learn how to determine trigonometric ratios for angles R and P in a right triangle using the Pythagorean theorem. This problem covers key concepts such as sine, cosine, and tangent ratios, along with a detailed step-by-step solution suitable for students in Grades 8-10.

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