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 will follow these steps:

### Step 1: Understand the Given Information
We are given that the perimeter of triangle ABC is 6 times the arithmetic mean of the sines of its angles. We also know that side \( a = 1 \).

### Step 2: Use the Sine Rule
According to the sine rule, we have:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \]
From this, we can express the sines of the angles in terms of \( k \):
\[ \sin A = ak, \quad \sin B = bk, \quad \sin C = ck \]

### Step 3: Calculate the Perimeter
The perimeter \( P \) of triangle ABC is given by:
\[ P = a + b + c \]

### Step 4: Calculate the Arithmetic Mean of the Sines
The arithmetic mean of the sines of the angles is:
\[ \text{Arithmetic Mean} = \frac{\sin A + \sin B + \sin C}{3} = \frac{ak + bk + ck}{3} = \frac{k(a + b + c)}{3} \]

### Step 5: Relate the Perimeter to the Arithmetic Mean
According to the problem, we have:
\[ P = 6 \times \text{Arithmetic Mean} \]
Substituting the expressions we derived:
\[ a + b + c = 6 \times \frac{k(a + b + c)}{3} \]
This simplifies to:
\[ a + b + c = 2k(a + b + c) \]

### Step 6: Solve for \( k \)
Assuming \( a + b + c \neq 0 \), we can divide both sides by \( a + b + c \):
\[ 1 = 2k \]
Thus, we find:
\[ k = \frac{1}{2} \]

### Step 7: Substitute \( k \) Back to Find \( \sin A \)
We know from the sine rule:
\[ \sin A = ak \]
Substituting \( a = 1 \) and \( k = \frac{1}{2} \):
\[ \sin A = 1 \cdot \frac{1}{2} = \frac{1}{2} \]

### Step 8: Find Angle \( A \)
To find angle \( A \), we take the inverse sine:
\[ A = \sin^{-1}\left(\frac{1}{2}\right) \]
This gives:
\[ A = 30^\circ \]

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

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