In `triangleABC, angleA = 45^(@) , angleB = 30^(@) , and b=8`. Side a =

Options

A. 6.5
B. 11
C. 12
D. 14

Explanation

To find the length of side \( a \) in triangle \( ABC \) where \( \angle A = 45^\circ \), \( \angle B = 30^\circ \), and side \( b = 8 \), we can use the Law of Sines.

### Step-by-Step Solution:

1. **Identify the angles and sides**:
- Given: \( \angle A = 45^\circ \), \( \angle B = 30^\circ \), and \( b = 8 \).
- We need to find \( a \).

2. **Calculate angle C**:
- Since the sum of angles in a triangle is \( 180^\circ \):
\[ \angle C = 180^\circ - \angle A - \angle B = 180^\circ - 45^\circ - 30^\circ = 105^\circ \]

3. **Apply the Law of Sines**:
- The Law of Sines states:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
- Substituting the known values:
\[ \frac{a}{\sin 45^\circ} = \frac{8}{\sin 30^\circ} \]

4. **Substitute the values of sine**:
- We know:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin 30^\circ = \frac{1}{2} \]
- Therefore, substituting these values gives:
\[ \frac{a}{\frac{1}{\sqrt{2}}} = \frac{8}{\frac{1}{2}} \]

5. **Simplify the right side**:
- The right side simplifies to:
\[ \frac{8}{\frac{1}{2}} = 8 \times 2 = 16 \]
- Now we have:
\[ \frac{a}{\frac{1}{\sqrt{2}}} = 16 \]

6. **Cross-multiply to solve for \( a \)**:
- Cross-multiplying gives:
\[ a = 16 \times \frac{1}{\sqrt{2}} = \frac{16}{\sqrt{2}} \]

7. **Rationalize the denominator**:
- To rationalize:
\[ a = \frac{16 \sqrt{2}}{2} = 8\sqrt{2} \]

8. **Approximate the value**:
- Using \( \sqrt{2} \approx 1.414 \):
\[ a \approx 8 \times 1.414 \approx 11.312 \]

### Final Answer:
Thus, the length of side \( a \) is approximately \( 11.31 \).