The perimeter of a `DeltaABC` is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then the angle A is

Options

A. `pi/6`
B. `pi/3`
C. `pi/2`
D. `pi`

Explanation

To solve the problem, we need to find the angle A of triangle ABC given that the perimeter of the triangle is 6 times the arithmetic mean of the sines of its angles, and that side a = 1.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The perimeter of triangle ABC is given by \( P = a + b + c \).
- The arithmetic mean of the sines of the angles is given by \( AM = \frac{\sin A + \sin B + \sin C}{3} \).
- We know that \( P = 6 \times AM \).

2. **Setting Up the Equation:**
- From the information above, we can write:
\[ a + b + c = 6 \times \frac{\sin A + \sin B + \sin C}{3} \]
- Simplifying this, we have:
\[ a + b + c = 2(\sin A + \sin B + \sin C) \]

3. **Using the Law of Sines:**
- According to the Law of Sines, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
- Therefore, we can express the sides in terms of k:
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

4. **Substituting into the Perimeter Equation:**
- Substituting the expressions for b and c into the perimeter equation:
\[ k \sin A + k \sin B + k \sin C = 2(\sin A + \sin B + \sin C) \]
- Factoring out \( \sin A + \sin B + \sin C \):
\[ k(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C) \]

5. **Solving for k:**
- Assuming \( \sin A + \sin B + \sin C \neq 0 \), we can divide both sides by \( \sin A + \sin B + \sin C \):
\[ k = 2 \]

6. **Finding the Sine of Angle A:**
- Since \( a = 1 \), we can substitute into the equation:
\[ a = k \sin A \implies 1 = 2 \sin A \]
- Therefore, we have:
\[ \sin A = \frac{1}{2} \]

7. **Determining Angle A:**
- The angle A for which \( \sin A = \frac{1}{2} \) is:
\[ A = 30^\circ \]

### Final Answer:
The angle A is \( 30^\circ \).