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To solve for \( x \) in the given triangle, we can use trigonometric functions since it’s a right triangle.

### Step-by-Step Solution

1. **Identify the sides and angle**:
   - The hypotenuse of the triangle is given as \( 19 \).
   - The angle opposite side \( x \) is \( 43^\circ \).
   - We need to find \( x \), which is the side opposite the \( 43^\circ \) angle.

2. **Choose the appropriate trigonometric function**:
   - Since we know the hypotenuse and need to find the opposite side, we use the sine function:
     \[
     \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
     \]
   - Here, \( \theta = 43^\circ \) and the hypotenuse is \( 19 \).

3. **Set up the equation**:
   \[
   \sin(43^\circ) = \frac{x}{19}
   \]

4. **Solve for \( x \)**:
   - Multiply both sides by \( 19 \):
     \[
     x = 19 \cdot \sin(43^\circ)
     \]

5. **Calculate the value**:
   - Using a calculator, \(\sin(43^\circ) \approx 0.682\).
   \[
   x \approx 19 \cdot 0.682 = 12.958
   \]
   - Rounding to the nearest tenth, \( x \approx 13.0 \).

### Final Answer
\[
x = 13.0
\]

Would you like more details or have any questions?

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Here are five additional questions related to this problem:

1. How would we solve for the length of the side adjacent to the \( 43^\circ \) angle?
2. What if we were asked to find the angle opposite side \( x \) given only \( x \) and the hypotenuse?
3. How could we verify the answer using the Pythagorean theorem?
4. What would change if the angle was \( 47^\circ \) instead of \( 43^\circ \)?
5. How would the solution differ if only the angle and one leg of the triangle were known?

**Tip**: Always make sure your calculator is in degree mode when solving trigonometry problems involving angles given in degrees.

Solve for x in the triangle. Round your answer to the nearest tenth.

Learn how to solve for the length of a side in a right triangle using the sine function. In this problem, we find the side opposite a 43-degree angle with a hypotenuse of 19, rounding the answer to the nearest tenth. Ideal for high school trigonometry students.

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