In an isosceles triangle ABC.
AB = BC = 10 cm and BC = 18 cm .
Find the value of :
` sin ^(2) B + cos ^(2) C `

Answer

To solve the problem, we need to find the value of \( \sin^2 B + \cos^2 C \) in the isosceles triangle ABC, where \( AB = AC = 10 \, \text{cm} \) and \( BC = 18 \, \text{cm} \).

### Step-by-step Solution:

1. **Identify the triangle and its properties**:
Given that \( AB = AC = 10 \, \text{cm} \) and \( BC = 18 \, \text{cm} \), we can label the triangle as follows:
- \( AB = AC = 10 \, \text{cm} \) (the equal sides)
- \( BC = 18 \, \text{cm} \) (the base)

2. **Determine angles B and C**:
Since \( AB = AC \), angles \( B \) and \( C \) are equal. Therefore, \( \angle B = \angle C \).

3. **Use the cosine rule to find cos B**:
The cosine rule states:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
Here, we can set:
- \( a = 10 \) (AB)
- \( b = 10 \) (AC)
- \( c = 18 \) (BC)

Plugging in the values:
\[ 18^2 = 10^2 + 10^2 - 2 \cdot 10 \cdot 10 \cdot \cos(B) \]
\[ 324 = 100 + 100 - 200 \cos(B) \]
\[ 324 = 200 - 200 \cos(B) \]
Rearranging gives:
\[ 200 \cos(B) = 200 - 324 \]
\[ 200 \cos(B) = -124 \]
\[ \cos(B) = -\frac{124}{200} = -\frac{31}{50} \]

4. **Find sin^2 B**:
Using the Pythagorean identity:
\[ \sin^2 B + \cos^2 B = 1 \]
We can find \( \sin^2 B \):
\[ \sin^2 B = 1 - \cos^2 B \]
\[ \cos^2 B = \left(-\frac{31}{50}\right)^2 = \frac{961}{2500} \]
Thus:
\[ \sin^2 B = 1 - \frac{961}{2500} = \frac{2500 - 961}{2500} = \frac{1539}{2500} \]

5. **Find cos^2 C**:
Since \( \angle B = \angle C \), we have:
\[ \cos C = \cos B = -\frac{31}{50} \]
Therefore:
\[ \cos^2 C = \left(-\frac{31}{50}\right)^2 = \frac{961}{2500} \]

6. **Calculate \( \sin^2 B + \cos^2 C \)**:
Now, we can find:
\[ \sin^2 B + \cos^2 C = \frac{1539}{2500} + \frac{961}{2500} = \frac{1539 + 961}{2500} = \frac{2500}{2500} = 1 \]

### Final Answer:
\[ \sin^2 B + \cos^2 C = 1 \]