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.

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.