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To solve for \( x \) in the given right triangle \( \triangle NMO \), we will use trigonometric relationships. Here's how we proceed:

### Step-by-Step Process:
1. **Identify the triangle sides and angle:**
   - \( \angle NMO = 52^\circ \) (the given angle).
   - \( NO = 7 \) (the side opposite the angle \( \angle NMO \)).
   - \( x = NM \) (the hypotenuse).

2. **Select the trigonometric function:**
   Since we have the opposite side (\( NO \)) and are looking for the hypotenuse (\( NM \)), we will use the **sine function**, which relates the opposite side and the hypotenuse.

   The sine function is given by:
   \[
   \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
   \]
   Substituting the known values:
   \[
   \sin(52^\circ) = \frac{7}{x}
   \]

3. **Solve for \( x \):**
   \[
   x = \frac{7}{\sin(52^\circ)}
   \]
   Using a calculator to find \( \sin(52^\circ) \):
   \[
   \sin(52^\circ) \approx 0.7880
   \]
   Therefore:
   \[
   x = \frac{7}{0.7880} \approx 8.9
   \]

So, the value of \( x \) is approximately **8.9**.

Would you like more details on any step or have further questions? 

Here are 5 related questions to explore:

1. How do you determine which trigonometric function to use based on the sides of a triangle?
2. Can you use other trigonometric functions (cosine, tangent) to solve for different unknowns in this triangle?
3. How would you calculate the length of side \( MO \) using trigonometry?
4. How do you check if a triangle is right-angled using side lengths?
5. How would the solution change if the angle given were different?

**Tip:** When solving trigonometric problems, always ensure your calculator is set to the correct mode (degrees or radians) depending on the angle given.

Solve for x. Round to the nearest tenth, if necessary. (Right triangle with angle 52°, opposite side length 7, and hypotenuse x).

Learn how to solve for the hypotenuse in a right triangle using the sine function. Given a 52° angle and an opposite side of length 7, this solution covers key trigonometric concepts and provides a step-by-step explanation. Suitable for students in Grades 9-12.

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