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To find the angle \( \angle B \) in triangle \( ABC \), where \( AB = 5 \, \text{cm} \), \( AC = 9 \, \text{cm} \), and \( BC = 10 \, \text{cm} \), we can use the **Cosine Rule**:

\[
\cos B = \frac{a^2 + c^2 - b^2}{2ac}
\]

where:
- \( a = 9 \, \text{cm} \) (side opposite \( B \)),
- \( b = 10 \, \text{cm} \) (side opposite \( A \)),
- \( c = 5 \, \text{cm} \) (side opposite \( C \)).

### Step-by-Step Calculation:

1. Substitute the values into the formula:

   \[
   \cos B = \frac{9^2 + 5^2 - 10^2}{2 \times 9 \times 5}
   \]

2. Calculate each term:
   - \( 9^2 = 81 \)
   - \( 5^2 = 25 \)
   - \( 10^2 = 100 \)

   So,

   \[
   \cos B = \frac{81 + 25 - 100}{2 \times 9 \times 5} = \frac{6}{90}
   \]

3. Simplify:

   \[
   \cos B = \frac{6}{90} = \frac{1}{15}
   \]

4. Find \( B \) by taking the inverse cosine:

   \[
   B = \cos^{-1}\left(\frac{1}{15}\right)
   \]

Using a calculator, \( B \approx 86.2^\circ \).

### Answer:
\[
\angle B \approx 86.2^\circ
\]

Would you like further details on any part of the calculation?

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### Related Questions:
1. How would you find angle \( C \) in this triangle?
2. Can this triangle be classified as a right triangle?
3. How would you use the Sine Rule to find the angles of this triangle?
4. What is the perimeter of triangle \( ABC \)?
5. How would you find the area of triangle \( ABC \)?

### Tip:
In triangles, the Cosine Rule is particularly useful for finding an angle when all three sides are known.

Learn how to find angle B in triangle ABC with side lengths AB = 5 cm, AC = 9 cm, and BC = 10 cm using the Cosine Rule. This trigonometry problem is ideal for students in Grades 9-12 and involves applying the formula cos B = (a^2 + c^2 - b^2) / (2ac).

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