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Express the trig ratios as fractions in simplest terms.

We are given a right triangle with side lengths:

- \( VW = 2 \) (one leg),
- \( WX = 3 \) (the other leg),
- \( VX = \sqrt{5} \) (the hypotenuse).

To solve the trigonometric ratios, we will use the following definitions:

1. **Sine** of an angle = \( \frac{\text{opposite}}{\text{hypotenuse}} \)
2. **Cosine** of an angle = \( \frac{\text{adjacent}}{\text{hypotenuse}} \)

### For \( \sin X \):
- The opposite side to \( X \) is \( VW = 2 \),
- The hypotenuse is \( VX = \sqrt{5} \).

Thus, 
\[
\sin X = \frac{VW}{VX} = \frac{2}{\sqrt{5}}.
\]
To rationalize the denominator, we multiply by \( \frac{\sqrt{5}}{\sqrt{5}} \):
\[
\sin X = \frac{2 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{2\sqrt{5}}{5}.
\]

### For \( \cos W \):
- The adjacent side to \( W \) is \( VW = 2 \),
- The hypotenuse is \( VX = \sqrt{5} \).

Thus, 
\[
\cos W = \frac{VW}{VX} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}.
\]

Therefore:

- \( \sin X = \frac{2\sqrt{5}}{5} \),
- \( \cos W = \frac{2\sqrt{5}}{5} \).

Would you like a breakdown of these steps or further questions on trigonometry?

Here are 5 related questions for further understanding:
1. How can you find \( \cos X \) and \( \sin W \) for the same triangle?
2. What is the tangent of \( X \) in this triangle?
3. How do you rationalize a denominator involving square roots?
4. Can you derive the Pythagorean identity using this triangle?
5. What would change in these calculations if the hypotenuse was a whole number?

**Tip:** Rationalizing square roots in the denominator simplifies expressions for easier comparison and interpretation.

Express the trig ratios as fractions in simplest terms for the triangle with side lengths 2, 3, and hypotenuse sqrt(5).

Solve for sine and cosine in a right triangle with side lengths 2, 3, and hypotenuse sqrt(5). This problem includes trigonometric concepts like sine, cosine, and rationalizing denominators. Suitable for students in Grades 9-10.

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