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This is a right triangle problem, where we need to solve for the side \( x \) using trigonometric functions. Here’s a breakdown of the steps:

1. **Given:**
   - Angle \( \angle ECD = 62^\circ \) (angle adjacent to side CD).
   - \( CD = 15 \) (the side adjacent to the angle).
   - \( x = ED \) (the opposite side).

2. **Trigonometric Function:**
   Since we know the angle and the adjacent side, and we need to find the opposite side, we use the **tangent** function:
   \[
   \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
   \]
   In this case:
   \[
   \tan(62^\circ) = \frac{x}{15}
   \]

3. **Solving for \( x \):**
   \[
   x = 15 \times \tan(62^\circ)
   \]

I will calculate this now.The length of side \( x \) (opposite to the given angle) is approximately 28.2 when rounded to the nearest tenth.

Would you like more details or further clarification on any steps?

Here are 5 related questions for further practice:
1. How would you calculate the hypotenuse of this triangle?
2. What if the angle was \( 45^\circ \)? How would that affect \( x \)?
3. Can you use the sine or cosine functions to solve for \( x \)?
4. How can you verify the answer using the Pythagorean theorem?
5. What is the measure of the third angle in this triangle?

**Tip:** Always keep track of which trigonometric function to use based on the sides involved (opposite, adjacent, hypotenuse).

Solve for x in the right triangle where angle ECD = 62°, side CD = 15 units, and side ED = x units. Round to the nearest tenth, if necessary.

Learn how to solve for the unknown side x in a right triangle where the angle is 62° and the adjacent side is 15 units. The solution involves applying the tangent function. Ideal for students in Grades 9-10.

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