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

Explanation

To solve the problem, we will use the given relationships between the angles and sides of triangle ABC:

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

2. **Using the Law of Sines**:
According to the Law of Sines, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
Substituting \( a = 2b \) and \( A = 3B \) into the equation gives us:
\[ \frac{2b}{\sin(3B)} = \frac{b}{\sin B} \]

3. **Simplifying the Equation**:
We can cancel \( b \) from both sides (assuming \( b \neq 0 \)):
\[ \frac{2}{\sin(3B)} = \frac{1}{\sin B} \]
Cross-multiplying results in:
\[ 2 \sin B = \sin(3B) \]

4. **Using the Sine Triple Angle Formula**:
The sine of a triple angle can be expressed as:
\[ \sin(3B) = 3\sin B - 4\sin^3 B \]
Substituting this into our equation gives:
\[ 2 \sin B = 3 \sin B - 4 \sin^3 B \]

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

6. **Factoring the Equation**:
Factoring out \( \sin B \):
\[ \sin B (1 - 4 \sin^2 B) = 0 \]
This gives us two cases:
- \( \sin B = 0 \)
- \( 1 - 4 \sin^2 B = 0 \)

7. **Solving Each Case**:
- For \( \sin B = 0 \), \( B = 0^\circ \) (not possible in a triangle).
- For \( 1 - 4 \sin^2 B = 0 \):
\[ 4 \sin^2 B = 1 \implies \sin^2 B = \frac{1}{4} \implies \sin B = \frac{1}{2} \]
Thus, \( B = 30^\circ \).

8. **Finding Angle A**:
Using \( A = 3B \):
\[ A = 3 \times 30^\circ = 90^\circ \]

9. **Conclusion**:
Since angle \( A = 90^\circ \), triangle ABC is a right triangle.

Thus, the correct answer is that triangle ABC is a right angle triangle.