1. Express the perimeter in terms of the sides \( a \), \( b \), and \( c \).
2. Express the arithmetic mean of the sines of the angles.
3. Use the given relationship to set up an equation.
4. Use the Law of Sines to relate the sides and angles.
5. Solve for angle \( A \).
Step 1: Express the Perimeter
The perimeter \( P \) of \( \Delta ABC \) is: \[ P = a + b + c \] Given that \( a = 1 \), so: \[ P = 1 + b + c \] Step 2: Express the Arithmetic Mean of the Sines of the Angles
The arithmetic mean of \( \sin A \), \( \sin B \), and \( \sin C \) is:
\[ \text{Mean} = \frac{\sin A + \sin B + \sin C}{3} \] Step 3: Use the Given Relationship
According to the problem: \[ P = 6 \times \text{Mean} \] Substituting the expressions we have: \[ 1 + b + c = 6 \times \left( \frac{\sin A + \sin B + \sin C}{3} \right ) \] Simplify the right side: \[ 1 + b + c = 2 (\sin A + \sin B + \sin C) \] So, we have: \[ 1 + b + c = 2 (\sin A + \sin B + \sin C) \quad \text{(Equation 1)} \] Step 4: Use the Law of Sines
The Law of Sines states: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \] where \( R \) is the radius of the circumscribed circle.
Given \( a = 1 \), we have: \[ \frac{1}{\sin A} = 2R \implies R = \frac{1}{2 \sin A} \] Similarly: \[ b = 2R \sin B = \frac{\sin B}{\sin A} \] \[ c = 2R \sin C = \frac{\sin C}{\sin A} \] Step 5: Substitute \( b \) and \( c \) into Equation 1 From Equation 1: \[ 1 + b + c = 2 (\sin A + \sin B + \sin C) \] Substitute \( b \) and \( c \): \[ 1 + \frac{\sin B}{\sin A} + \frac{\sin C}{\sin A} = 2 (\sin A + \sin B + \sin C) \] Combine the terms on the left: \[ 1 + \frac{\sin B + \sin C}{\sin A} = 2 (\sin A + \sin B + \sin C) \] Step 6: Simplify the Equation Let's denote \( S = \sin A + \sin B + \sin C \). Then the equation becomes: \[ 1 + \frac{S - \sin A}{\sin A} = 2S \] Simplify the left side: \[ 1 + \frac{S}{\sin A} - 1 = 2S \] The \( 1 \) and \( -1 \) cancel out: \[ \frac{S}{\sin A} = 2S \] Assuming \( S \neq 0 \) (which it shouldn't be, since the sum of sines of angles in a triangle is positive), we can divide both sides by \( S \): \[ \frac{1}{\sin A} = 2 \] Step 7: Solve for \( \sin A \) From the equation: \[ \frac{1}{\sin A} = 2 \implies \sin A = \frac{1}{2} \] Step 8: Determine Angle \( A \) We know that \( \sin A = \frac{1}{2} \). The angles in a triangle are between \( 0 \) and \( \pi \) radians (or \( 0^\circ \) and \( 180^\circ \)). The sine function equals \( \frac{1}{2} \) at: \[ A = 30^\circ \quad \text{or} \quad A = 150^\circ \] However, in a triangle, the sum of angles is \( 180^\circ \). If \( A = 150^\circ \), then \( B + C = 30^\circ \), which would make \( \sin B + \sin C \) very small, potentially leading to a contradiction with the earlier equation. Therefore, the more plausible solution is: \[ A = 30^\circ \] Verification:
Let's verify if \( A = 30^\circ \) satisfies all given conditions.
Given \( A = 30^\circ \), \( \sin A = \frac{1}{2} \).
Using the Law of Sines:
\[ \frac{a}{\sin A} = \frac{1}{\frac{1}{2}} = 2 \] So, \[ \frac{b}{\sin B} = 2 \implies b = 2 \sin B \] \[ \frac{c}{\sin C} = 2 \implies c = 2 \sin C \] The perimeter: \[ P = a + b + c = 1 + 2 \sin B + 2 \sin C \] The arithmetic mean of the sines: \[ \text{Mean} = \frac{\sin A + \sin B + \sin C}{3} = \frac{\frac{1}{2} + \sin B + \sin C}{3} \] According to the problem: \[ P = 6 \times \text{Mean} \] Substitute \( P \) and \( \text{Mean} \): \[ 1 + 2 \sin B + 2 \sin C = 6 \times \left( \frac{\frac{1}{2} + \sin B + \sin C}{3} \right ) \] Simplify the right side: \[ 1 + 2 \sin B + 2 \sin C = 2 \left( \frac{1}{2} + \sin B + \sin C \right ) = 1 + 2 \sin B + 2 \sin C \] Both sides are equal, so our solution \( A = 30^\circ \) satisfies the given conditions.
Final Answer:
The measure of angle \( A \) is:
\[ {\dfrac{\pi}{6}} \]