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.