In `triangleABC`, with usual notations b=2c and B=3C, then sinA=

Options

A. `sqrt(3)/2`
B. 1
C. `1/2`
D. `1/sqrt(2)`

Answer

To solve the problem step by step, we start with the given information about triangle \( ABC \):

1. **Given Information**:
- \( b = 2c \) (where \( b \) is the side opposite angle \( B \) and \( c \) is the side opposite angle \( C \))
- \( B = 3C \)

2. **Using the Sine Rule**:
The sine rule states that:
\[ \frac{b}{\sin B} = \frac{c}{\sin C} \]
Substituting the values of \( b \) and \( B \):
\[ \frac{2c}{\sin(3C)} = \frac{c}{\sin C} \]

3. **Cancelling \( c \)**:
Since \( c \) is not zero, we can cancel \( c \) from both sides:
\[ \frac{2}{\sin(3C)} = \frac{1}{\sin C} \]

4. **Cross Multiplying**:
Cross multiplying gives:
\[ 2 \sin C = \sin(3C) \]

5. **Using the Sine Triple Angle Formula**:
The formula for \( \sin(3\theta) \) is:
\[ \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \]
Applying this to our equation:
\[ 2 \sin C = 3 \sin C - 4 \sin^3 C \]

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

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

Solving \( 4 \sin^2 C - 1 = 0 \):
\[ 4 \sin^2 C = 1 \implies \sin^2 C = \frac{1}{4} \implies \sin C = \frac{1}{2} \]

8. **Finding Angle \( C \)**:
Since \( \sin C = \frac{1}{2} \), we have:
\[ C = 30^\circ \]

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

10. **Finding Angle \( A \)**:
The sum of angles in a triangle is \( 180^\circ \):
\[ A + B + C = 180^\circ \implies A + 90^\circ + 30^\circ = 180^\circ \implies A = 60^\circ \]

11. **Finding \( \sin A \)**:
Finally, we need to find \( \sin A \):
\[ \sin A = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

Thus, the final answer is:
\[ \sin A = \frac{\sqrt{3}}{2} \]