In a `DeltaABC`, a = 2, b = 3 and sinA = `2/3`, find `angleB`.

Answer

To solve the problem, we will use the Law of Sines, which states that the ratio of the sine of an angle to the length of its opposite side is constant for all angles in a triangle.

### Step-by-Step Solution:

1. **Identify the given values**:
- Side \( a = 2 \)
- Side \( b = 3 \)
- \( \sin A = \frac{2}{3} \)

2. **Use the Law of Sines**:
According to the Law of Sines:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} \]
Substituting the known values:
\[ \frac{\frac{2}{3}}{2} = \frac{\sin B}{3} \]

3. **Simplify the left side**:
\[ \frac{2}{3} \div 2 = \frac{2}{3} \times \frac{1}{2} = \frac{2}{6} = \frac{1}{3} \]
So, we have:
\[ \frac{1}{3} = \frac{\sin B}{3} \]

4. **Cross-multiply to solve for \( \sin B \)**:
\[ 1 \cdot 3 = \sin B \cdot 3 \]
Thus:
\[ \sin B = 1 \]

5. **Find angle \( B \)**:
The sine of an angle is equal to 1 when the angle is \( 90^\circ \):
\[ B = 90^\circ \]

### Final Answer:
Angle \( B \) is \( 90^\circ \).