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The problem provided is a right triangle \( \triangle PON \), where:

- \( \angle PON = 90^\circ \) (a right angle),
- \( \angle P = 60^\circ \),
- \( PN = 9 \) (the hypotenuse of the triangle),
- You are asked to solve for \( x \), which represents the length of the side \( PO \).

### Approach:

Since we are working with a right triangle and we are given one angle \( 60^\circ \) and the hypotenuse (9), we can use trigonometric ratios to solve for \( x = PO \). The cosine function relates the adjacent side to the hypotenuse in a right triangle:

\[
\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}
\]

Here:
- \( \theta = 60^\circ \),
- adjacent = \( x = PO \),
- hypotenuse = 9.

So:

\[
\cos(60^\circ) = \frac{x}{9}
\]

Using \( \cos(60^\circ) = 0.5 \), we substitute this into the equation:

\[
0.5 = \frac{x}{9}
\]

Now, solve for \( x \):

\[
x = 0.5 \times 9 = 4.5
\]

Thus, the length of side \( PO \) is \( \boxed{4.5} \) units.

Would you like more details or further clarification on this problem?

Here are 5 related questions for further exploration:
1. How would you solve for the length of side \( ON \) using trigonometry?
2. What trigonometric ratio relates the opposite side to the hypotenuse in this triangle?
3. If the hypotenuse were 12 instead of 9, what would \( x \) be?
4. Can you explain how the Pythagorean theorem could be applied here after solving for \( x \)?
5. How would the solution change if \( \angle P \) were \( 45^\circ \) instead of \( 60^\circ \)?

**Tip:** In any right triangle, remember that the cosine function relates the adjacent side to the hypotenuse, and the sine function relates the opposite side to the hypotenuse.

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

Learn how to solve for the side length of a right triangle using the cosine function. This problem involves a 60° angle and a hypotenuse of 9 units. Suitable for students in Grades 9-11, this example demonstrates trigonometric applications.

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