Find the circumradius of `Delta ABC` in which `angle A=45^(@)` side `a=8sqrt2`.

Answer

To find the circumradius \( R \) of triangle \( ABC \) where \( \angle A = 45^\circ \) and side \( a = 8\sqrt{2} \), we can follow these steps:

### Step 1: Identify the triangle type
Since we have \( \angle A = 45^\circ \) and we are given one side, we can assume that triangle \( ABC \) is a right triangle with \( \angle B = 90^\circ \) and \( \angle C = 45^\circ \). This is because the angles in a triangle sum up to \( 180^\circ \).

### Step 2: Use the circumradius formula for a right triangle
For a right triangle, the circumradius \( R \) can be calculated using the formula:
\[ R = \frac{a}{2 \sin A} \]
where \( a \) is the side opposite to angle \( A \) and \( A \) is the angle at vertex \( A \).

### Step 3: Substitute the known values
We know:
- \( a = 8\sqrt{2} \)
- \( A = 45^\circ \)

Now, we can substitute these values into the formula:
\[ R = \frac{8\sqrt{2}}{2 \sin 45^\circ} \]

### Step 4: Calculate \( \sin 45^\circ \)
The value of \( \sin 45^\circ \) is:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]

### Step 5: Substitute \( \sin 45^\circ \) into the formula
Now substituting \( \sin 45^\circ \) into the equation for \( R \):
\[ R = \frac{8\sqrt{2}}{2 \cdot \frac{1}{\sqrt{2}}} \]

### Step 6: Simplify the expression
Now simplify the expression:
\[ R = \frac{8\sqrt{2}}{2 \cdot \frac{1}{\sqrt{2}}} = \frac{8\sqrt{2} \cdot \sqrt{2}}{2} = \frac{8 \cdot 2}{2} = \frac{16}{2} = 8 \]

### Conclusion
Thus, the circumradius \( R \) of triangle \( ABC \) is:
\[ \boxed{8} \]