The perimeter of atriangle ABC is 6 times the arihmetic mean of the sines of its angles .If the side a is 1. the `angle A` is

Options

A. `30^@`
B. `60 ^@`
C. `90^@`
D. `120^@`

Answer

To solve the problem, we need to find the angle A in 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 \( A, B, C \) is given by:
\[ \text{Arithmetic Mean} = \frac{\sin A + \sin B + \sin C}{3} \]
- We are told that:
\[ P = 6 \times \text{Arithmetic Mean} \]

2. **Using the Sine Rule:**
- According to the sine rule:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \]
- Therefore, we can express the sines of the angles as:
\[ \sin A = ak, \quad \sin B = bk, \quad \sin C = ck \]

3. **Substituting into the Perimeter Equation:**
- The perimeter can be expressed as:
\[ P = a + b + c \]
- Substituting the values of \( \sin A, \sin B, \sin C \) into the arithmetic mean:
\[ \text{Arithmetic Mean} = \frac{ak + bk + ck}{3} = \frac{k(a + b + c)}{3} \]

4. **Setting Up the Equation:**
- From the problem statement, we have:
\[ a + b + c = 6 \times \frac{k(a + b + c)}{3} \]
- Simplifying this gives:
\[ a + b + c = 2k(a + b + c) \]
- Assuming \( a + b + c \neq 0 \), we can divide both sides by \( a + b + c \):
\[ 1 = 2k \implies k = \frac{1}{2} \]

5. **Finding the Sine Values:**
- Now substituting \( k \) back into the sine equations:
\[ \sin A = ak = 1 \cdot \frac{1}{2} = \frac{1}{2} \]
- Therefore, we find:
\[ \sin A = \frac{1}{2} \]

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

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