In a triangle ABC, if a = 2b and A = 3B then which one of the following is correct?

Options

A. The triangle is obtuse - angled
B. The triangle is acute -angled but not right - angled
C. The triangle is right - angled
D. The triangle is isosceles but not obtuse - angled

Answer

To solve the problem step by step, we will use the given relationships and apply the sine law in triangle ABC.

### Step-by-Step Solution:

1. **Given Relationships**:
- We have \( a = 2b \) and \( A = 3B \).

2. **Applying the Sine Law**:
- The sine law states that:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} \]
- Substituting the values we have:
\[ \frac{\sin(3B)}{2b} = \frac{\sin B}{b} \]

3. **Cross-Multiplying**:
- Cross-multiplying gives us:
\[ \sin(3B) = 2 \sin B \]

4. **Using the Sine Triple Angle Formula**:
- The sine triple angle formula states:
\[ \sin(3x) = 3\sin x - 4\sin^3 x \]
- Applying this to our equation:
\[ 3\sin B - 4\sin^3 B = 2\sin B \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 3\sin B - 2\sin B - 4\sin^3 B = 0 \]
- This simplifies to:
\[ \sin B - 4\sin^3 B = 0 \]

6. **Factoring Out \(\sin B\)**:
- Factoring out \(\sin B\):
\[ \sin B(1 - 4\sin^2 B) = 0 \]

7. **Finding Solutions**:
- This gives us two cases:
- \( \sin B = 0 \) (not valid in a triangle)
- \( 1 - 4\sin^2 B = 0 \)

8. **Solving for \(\sin B\)**:
- From \( 1 - 4\sin^2 B = 0 \):
\[ 4\sin^2 B = 1 \implies \sin^2 B = \frac{1}{4} \implies \sin B = \frac{1}{2} \]
- Therefore, \( B = 30^\circ \).

9. **Finding Angle A**:
- Since \( A = 3B \):
\[ A = 3 \times 30^\circ = 90^\circ \]

10. **Conclusion**:
- In triangle ABC, we have \( A = 90^\circ \) and \( B = 30^\circ \). The third angle \( C \) can be calculated as:
\[ C = 180^\circ - A - B = 180^\circ - 90^\circ - 30^\circ = 60^\circ \]
- Thus, the triangle is a right triangle.

### Answer:
The correct option is that the triangle is a right triangle.