Question

Express the trig ratios as fractions in simplest terms.
W
X
Y
24
51
45
Answer
Attempt 1 out of 2
The number of attempts on this specific problem before you are marked incorrect

sin YY =

cos XX =

sin YY and cos XX
.
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i know y = 24 and x= 51 but are they equal or unequal

Answer 1

Let's analyze the problem using the information given:

You have angles W, X, and Y, with corresponding opposite sides or heights. We are trying to find the sine and cosine ratios in simplest terms.

1. **Understanding the relationships**:
- **sin Y**: This is defined as the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle. You mentioned that the opposite side for angle Y is 24. We'll need to find the hypotenuse to express sin Y.
- **cos X**: This is defined as the ratio of the length of the adjacent side to the hypotenuse in a right triangle. You mentioned that the adjacent side for angle X is 51. Again, we will need the hypotenuse.

2. **Find the hypotenuse**:
Assuming that angles W, X, and Y relate to the sides 24 and 51, we can use the Pythagorean theorem to find the hypotenuse (h).

\[
h = \sqrt{(24^2) + (51^2)} = \sqrt{576 + 2601} = \sqrt{3177}
\]

Thus, the hypotenuse \( h \) simplifies to \( h = 3\sqrt{353} \).

3. **Calculating the trig ratios**:
- **sin Y**:
\[
\sin Y = \frac{\text{Opposite side to Y}}{\text{Hypotenuse}} = \frac{24}{\sqrt{3177}}
\]

To rationalize the denominator:
\[
\sin Y = \frac{24\sqrt{3177}}{3177}
\]

This fraction may need extra simplification or can be left in this form depending on your context.

- **cos X**:
\[
\cos X = \frac{\text{Adjacent side to X}}{\text{Hypotenuse}} = \frac{51}{\sqrt{3177}}
\]

Rationalizing, we get:
\[
\cos X = \frac{51\sqrt{3177}}{3177}
\]

So we have:
- \(\sin Y = \frac{24}{\sqrt{3177}}\)
- \(\cos X = \frac{51}{\sqrt{3177}}\)

Finally, they are not equal because they describe different trigonometric functions for different angles, and they depend on different opposite or adjacent sides in relation to their respective angles in the triangle.