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To solve for \( x \) in this triangle, we can use trigonometric ratios. The angle at \( H \) is given as \( 27^\circ \), and we have a right triangle \( \triangle HJI \) with the following details:

- \( \angle H = 27^\circ \)
- \( HI = 6 \) (the side adjacent to \( \angle H \))
- \( x = JI \) (the hypotenuse)

### Step-by-Step Solution

Since we are given the adjacent side (\( HI \)) and need to find the hypotenuse (\( JI \)), the appropriate trigonometric function to use is the cosine function:

\[
\cos(27^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{HI}{JI}
\]

Substitute the known values:

\[
\cos(27^\circ) = \frac{6}{x}
\]

Now, solve for \( x \):

1. Rearrange the equation to isolate \( x \):
   \[
   x = \frac{6}{\cos(27^\circ)}
   \]

2. Calculate \( \cos(27^\circ) \):
   \[
   \cos(27^\circ) \approx 0.8910
   \]

3. Substitute this value to find \( x \):
   \[
   x \approx \frac{6}{0.8910} \approx 6.7
   \]

### Answer
\( x \approx 6.7 \)

Let me know if you'd like further details or have any questions. 

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**Related Questions:**
1. What is the value of the opposite side (\( JH \)) using trigonometry?
2. How would the answer change if \( HI \) was a different length?
3. What trigonometric function would be used if we were finding \( JH \) instead?
4. How can you verify the answer using the Pythagorean theorem?
5. What is the sine of \( 27^\circ \) and how would it relate to finding other sides?

**Tip:** Always remember to double-check trigonometric values when rounding to ensure accuracy.

Solve for x. Round to the nearest tenth, if necessary.

Learn how to solve for the hypotenuse of a right triangle given an angle and one side. This problem involves using the cosine function, right triangle trigonometry, and rounding to the nearest tenth. Suitable for students in Grades 9-10.

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