Given the figure above, if ÐX = 35°, h = 6 cm, and mc027-1.jpg = 24 cm, what is the length of line w to the nearest tenth of a centimetre?  Assume line w meets line mc027-2.jpg at its midpoint.

To solve this problem, let's assume that the figure in question involves a right triangle or some geometric configuration where we can use trigonometric principles. Since we don't have the actual figure, we will work based on the given dataTo solve this problem, let's assume that the figure in question involves a right triangle or some geometric configuration where we can use trigonometric principles. Since we don't have the actual figure, we will work based on the given 

- \(\angle X = 35^\circ\)
- \(h = 6 \, \text{cm}\)
- \(mc027-1.jpg = 24 \, \text{cm}\)
- \(w\) meets \(mc027-2.jpg\) at its midpoint.

Let's assume the following configuration based on common problems of this type:
- \(w\) is a line segment drawn from the vertex opposite \(\angle X\) to the midpoint of the opposite side (let's call this point M).

In a triangle, the length of a line segment drawn from a vertex to the midpoint of the opposite side (a median) can be calculated using trigonometry and properties of triangles. Here, we will use some trigonometric relationships.

We can proceed using the Law of Cosines in the assumed configuration if necessary, but let's first consider the use of trigonometric ratios in the right triangle formed by the given dimensions.

If we assume that the triangle is divided into two smaller right triangles by the median, we can use trigonometric functions to solve for \(w\). 

First, let's consider the right triangle formed by the given dimensions:
1. \(\angle X = 35^\circ\)
2. Opposite side (height \(h\)) = 6 cm
3. Adjacent side (half of \(mc027-1.jpg\)) = 24 cm / 2 = 12 cm

Using the tangent function, we can find the length of the adjacent side:
\[
\tan(35^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{12}
\]

Calculating the adjacent side:
\[
\text{adjacent} = \frac{6}{\tan(35^\circ)}
\]

To find the hypotenuse (which could relate to \(w\)):
\[
\text{hypotenuse} = \sqrt{\text{opposite}^2 + \text{adjacent}^2} = \sqrt{6^2 + 12^2}
\]

Let's calculate these values:
\[
\tan(35^\circ) \approx 0.7002
\]
\[
\text{adjacent} = \frac{6}{0.7002} \approx 8.57 \, \text{cm}
\]
\[
\text{hypotenuse} = \sqrt{6^2 + 8.57^2} \approx \sqrt{36 + 73.45} \approx \sqrt{109.45} \approx 10.46 \, \text{cm}
\]

Therefore, the length of \(w\) (the hypotenuse in this case) is approximately 10.5 cm to the nearest tenth of a centimeter.

### Final Answer:
The length of line \(w\) is approximately **10.5 cm** to the nearest tenth of a centimeter.

Would you like more details on this solution, or do you have any other questions? Here are some further questions you might consider:

1. How would you solve a similar problem with different angle measurements?
2. What other trigonometric identities can be used to solve triangle problems?
3. How does the Pythagorean theorem apply in non-right triangles?
4. Can we use the Law of Sines to solve this problem?
5. How would the solution change if we had a different length for the hypotenuse?

### Tip:
Always double-check your angle measurements and ensure your calculator is in the correct mode (degrees or radians) when working with trigonometric functions.

Discover how to determine the length of line w in a triangle given that angle X = 35°, height h = 6 cm, and side mc027-1.jpg = 24 cm. The solution involves trigonometric ratios and properties of medians, providing a comprehensive understanding suitable for high school students.

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