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The given triangle shows a right triangle split into two smaller right triangles, with angles of 30° and 45°, and a perpendicular height of 44 units. We are asked to find the base length \( x \) (the side opposite the 45° angle).

### Step-by-step approach:

1. The larger triangle has angles of 30°, 45°, and 90°. The perpendicular creates two right triangles.
   
2. The side of length 44 forms the height of the larger triangle, and this height corresponds to the opposite side of the 30° angle in the left smaller triangle.

#### Using trigonometric relationships:
- In the left right triangle:
   \[
   \sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{44}{\text{hypotenuse}}
   \]
   Since \( \sin(30^\circ) = \frac{1}{2} \), we can solve for the hypotenuse (the side opposite the 45° angle in the larger triangle).

Let's calculate this value and use it to solve for \( x \).The hypotenuse of the left smaller triangle (which is the side opposite the 45° angle in the larger triangle) is 88 units.

Now, to find \( x \), the base of the larger triangle:

In the right triangle with a 45° angle (the larger triangle on the right side):
\[
\cos(45^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{88}
\]
Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \), we can solve for \( x \).

Let's calculate this value now.The base length \( x \) of the triangle is approximately 62.23 units.

Would you like any further details or have questions about this process? 

Here are five related questions you might consider:
1. How do trigonometric ratios like sine and cosine relate to the sides of a right triangle?
2. How can you solve a right triangle using other angles and sides?
3. What happens to the trigonometric values if the angle changes slightly?
4. How would the solution change if the angles were different (e.g., 60° and 30°)?
5. Can this method be applied to non-right triangles? If so, how?

**Tip**: When dealing with triangles, breaking them down into smaller right triangles can simplify complex problems!

In the given right triangle with angles 30°, 45°, and 90°, and a perpendicular height of 44 units, find the base length x (side opposite the 45° angle).

Solve for the base length in a right triangle with angles 30° and 45°, and a given height of 44 units. This problem uses trigonometric ratios and is suitable for students in Grades 9-11.

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