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To solve for \(x\), the missing side of the triangle, we can use the formula for the area of a triangle, which is given by:

\[
\text{Area} = \frac{1}{2}ab \sin C
\]

Where:
- \(a\) and \(b\) are the two sides of the triangle,
- \(C\) is the angle between them.

In this case:
- The area is \(26 \, \text{m}^2\),
- One side is \(13 \, \text{m}\),
- The angle \(C\) is \(30^\circ\),
- The other side is \(x\) (which we are solving for).

We can substitute the known values into the formula:

\[
26 = \frac{1}{2} \times 13 \times x \times \sin(30^\circ)
\]

Since \(\sin(30^\circ) = 0.5\), the equation becomes:

\[
26 = \frac{1}{2} \times 13 \times x \times 0.5
\]

Simplifying:

\[
26 = 3.25x
\]

Now, solve for \(x\):

\[
x = \frac{26}{3.25} = 8
\]

So, the length of \(x\) is \(8.0 \, \text{m}\).

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the solution change if the area was doubled?
2. What if the angle was \(45^\circ\) instead of \(30^\circ\)?
3. How would you calculate the perimeter of this triangle?
4. Can you derive the sine of other key angles like \(45^\circ\) and \(60^\circ\)?
5. How would this approach differ for a right-angled triangle?

**Tip**: Always check if you're given angles in degrees or radians before using trigonometric functions!

Work out the length x in the triangle with an area of 26 m², a base of 13 m, and an angle of 30 degrees.

Learn how to calculate the missing side of a triangle using trigonometric principles, given the area (26 m²), base (13 m), and angle (30°). This problem covers key geometry and trigonometry concepts suitable for Grades 10-12 students.

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