If in `DeltaABC`, a=2b and A=3B, then A is equal to

Options

A. `90^(@)`
B. `60^(@)`
C. `30^(@)`
D. `45^(@)`

Answer

To solve the problem where in triangle \( \Delta ABC \), \( a = 2b \) and \( A = 3B \), we need to find the value of angle \( A \).

### Step-by-Step Solution:

1. **Understand the Given Relationships**:
We are given that:
\[ a = 2b \quad \text{and} \quad A = 3B \]

2. **Apply the Sine Rule**:
According to the sine rule in a triangle:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
Substituting \( a = 2b \) into the sine rule gives:
\[ \frac{2b}{\sin A} = \frac{b}{\sin B} \]

3. **Simplify the Equation**:
We can cancel \( b \) (assuming \( b \neq 0 \)):
\[ \frac{2}{\sin A} = \frac{1}{\sin B} \]
Rearranging gives:
\[ 2 \sin B = \sin A \]

4. **Substitute \( A \) in Terms of \( B \)**:
Since \( A = 3B \), we substitute this into the equation:
\[ 2 \sin B = \sin(3B) \]

5. **Use the Sine Triple Angle Formula**:
The sine of triple angle can be expressed as:
\[ \sin(3B) = 3 \sin B - 4 \sin^3 B \]
Thus, we can rewrite our equation:
\[ 2 \sin B = 3 \sin B - 4 \sin^3 B \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 4 \sin^3 B - \sin B = 0 \]
Factoring out \( \sin B \):
\[ \sin B (4 \sin^2 B - 1) = 0 \]

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

Solving \( 4 \sin^2 B - 1 = 0 \):
\[ 4 \sin^2 B = 1 \implies \sin^2 B = \frac{1}{4} \implies \sin B = \frac{1}{2} \text{ or } \sin B = -\frac{1}{2} \]
Since angles in a triangle cannot be negative, we take:
\[ \sin B = \frac{1}{2} \]

8. **Determine the Angle \( B \)**:
The angle \( B \) corresponding to \( \sin B = \frac{1}{2} \) is:
\[ B = 30^\circ \]

9. **Calculate Angle \( A \)**:
Now, substituting back to find \( A \):
\[ A = 3B = 3 \times 30^\circ = 90^\circ \]

### Final Answer:
Thus, the value of angle \( A \) is:
\[ \boxed{90^\circ} \]