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The problem shows a triangle with sides 13 cm, 10 cm, and an unknown side \( x \). The angle between the two known sides is \( 100^\circ \).

To solve for the unknown side \( x \), we can use the **Cosine Rule**. The Cosine Rule is applicable for non-right triangles and is given by:

\[
x^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]

Where:
- \( a = 13 \, \text{cm} \)
- \( b = 10 \, \text{cm} \)
- \( C = 100^\circ \)

### Step-by-Step Solution:

1. Plug in the known values into the formula:

\[
x^2 = 13^2 + 10^2 - 2 \times 13 \times 10 \times \cos(100^\circ)
\]

2. Simplify the squares and multiplication:

\[
x^2 = 169 + 100 - 2 \times 13 \times 10 \times \cos(100^\circ)
\]
\[
x^2 = 269 - 260 \times \cos(100^\circ)
\]

3. Use a calculator to find \( \cos(100^\circ) \). The cosine of \( 100^\circ \) is approximately \( -0.1736 \).

\[
x^2 = 269 - 260 \times (-0.1736)
\]
\[
x^2 = 269 + 45.136
\]
\[
x^2 = 314.136
\]

4. Finally, take the square root of both sides to find \( x \):

\[
x = \sqrt{314.136} \approx 17.7 \, \text{cm}
\]

So, the value of \( x \) is approximately **17.7 cm**.

Would you like further details on this solution?

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Here are five follow-up questions related to this topic:

1. How does the Cosine Rule change if the given angle is \( 90^\circ \)?
2. Can you derive the Cosine Rule from the Pythagorean theorem?
3. What other conditions must be met to use the Sine Rule instead of the Cosine Rule?
4. How would you solve for an unknown angle if all three sides of the triangle were known?
5. What happens to the triangle if the angle between the two sides is less than \( 90^\circ \)?

**Tip:** The Cosine Rule is especially useful when dealing with non-right triangles where two sides and the included angle are known.

Work out the value of x in a triangle with sides 13 cm and 10 cm, and an included angle of 100°. Give your answer to 1 decimal place.

Learn how to solve for the unknown side of a triangle using the Cosine Rule, with sides measuring 13 cm and 10 cm and an angle of 100°. Detailed step-by-step solution provided, suitable for students in Grades 9-11.

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