In `Delta ABC`, a = 3, b = 4 and `sin A = (3)/(4)`, find `angle B`.

Answer

To find angle B in triangle ABC given that \( a = 3 \), \( b = 4 \), and \( \sin A = \frac{3}{4} \), we can use the Law of Sines.

### Step-by-Step Solution:

1. **Use the Law of Sines**:
The Law of Sines states that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
Here, we know \( a = 3 \), \( b = 4 \), and \( \sin A = \frac{3}{4} \).

2. **Substitute the known values into the equation**:
\[ \frac{3}{\frac{3}{4}} = \frac{4}{\sin B} \]

3. **Simplify the left side**:
\[ \frac{3}{\frac{3}{4}} = 3 \cdot \frac{4}{3} = 4 \]
So, we have:
\[ 4 = \frac{4}{\sin B} \]

4. **Cross-multiply to solve for \(\sin B\)**:
\[ 4 \sin B = 4 \]

5. **Divide both sides by 4**:
\[ \sin B = 1 \]

6. **Find angle B**:
The sine of an angle is equal to 1 at:
\[ B = \frac{\pi}{2} \text{ or } 90^\circ \]

### Final Answer:
\[ \text{Angle B} = 90^\circ \]