In a `DeltaABC`, if a = 5, `B=45^(@)` and `c=2sqrt(2)`, then b=

Options

A. `sqrt(3)`
B. 6
C. `2sqrt(13)`
D. `sqrt(13)`

Explanation

To solve for \( b \) in triangle \( ABC \) with given values \( a = 5 \), \( B = 45^\circ \), and \( c = 2\sqrt{2} \), we will use the Law of Cosines.

### Step-by-Step Solution:

1. **Write the Law of Cosines Formula**:
The Law of Cosines states that:
\[ b^2 = a^2 + c^2 - 2ac \cdot \cos(B) \]

2. **Substitute the Known Values**:
We know:
- \( a = 5 \)
- \( c = 2\sqrt{2} \)
- \( B = 45^\circ \)

Substitute these values into the formula:
\[ b^2 = 5^2 + (2\sqrt{2})^2 - 2 \cdot 5 \cdot 2\sqrt{2} \cdot \cos(45^\circ) \]

3. **Calculate Each Term**:
- \( 5^2 = 25 \)
- \( (2\sqrt{2})^2 = 4 \cdot 2 = 8 \)
- \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)

Now substitute these values:
\[ b^2 = 25 + 8 - 2 \cdot 5 \cdot 2\sqrt{2} \cdot \frac{1}{\sqrt{2}} \]

4. **Simplify the Expression**:
- The term \( 2 \cdot 5 \cdot 2\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \)

Thus, we have:
\[ b^2 = 25 + 8 - 20 \]
\[ b^2 = 13 \]

5. **Take the Square Root**:
To find \( b \), take the square root of both sides:
\[ b = \sqrt{13} \]

### Final Answer:
Thus, the value of \( b \) is \( \sqrt{13} \).

---