In a `Delta ABC, a = 8 sqrt(2), angle A = 95^(@)` . find circum radius

Answer

To find the circumradius \( R \) of triangle \( ABC \) given that \( a = 8\sqrt{2} \) and \( \angle A = 95^\circ \), we can use the formula derived from the sine rule:

\[ R = \frac{a}{2 \sin A} \]

### Step-by-Step Solution:

1. **Identify the known values**:
- Side \( a = 8\sqrt{2} \)
- Angle \( A = 95^\circ \)

2. **Calculate \( \sin A \)**:
- We need to find \( \sin(95^\circ) \). Using a calculator or sine table:
\[ \sin(95^\circ) \approx 0.9962 \]

3. **Substitute values into the circumradius formula**:
- Plugging the values into the formula \( R = \frac{a}{2 \sin A} \):
\[ R = \frac{8\sqrt{2}}{2 \cdot 0.9962} \]

4. **Simplify the expression**:
- First, calculate \( 2 \cdot 0.9962 \):
\[ 2 \cdot 0.9962 \approx 1.9924 \]
- Now, substitute this back into the formula:
\[ R = \frac{8\sqrt{2}}{1.9924} \]

5. **Calculate \( 8\sqrt{2} \)**:
- We know \( \sqrt{2} \approx 1.4142 \), thus:
\[ 8\sqrt{2} \approx 8 \cdot 1.4142 \approx 11.3136 \]

6. **Final calculation for \( R \)**:
- Now, substitute \( 8\sqrt{2} \) back into the equation:
\[ R \approx \frac{11.3136}{1.9924} \approx 5.68 \]

### Conclusion:
The circumradius \( R \) of triangle \( ABC \) is approximately \( 5.68 \).