Question:

**Step 1:** Define the perimeter and apply the Sine Rule. Let the sides of the triangle be $a, b, c$ opposite to angles $A, B, C$ respectively. The perimeter is $P = a + b + c$. By the Sine Rule, $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$ where $R$ is the circumradius. From this, we can express the sides in terms of $R$ and the sines of the angles: $$ a = 2R\sin A, \quad b = 2R\sin B, \quad c = 2R\sin C $$ **Step 2:** Express the perimeter in terms of $R$ and sines of angles. Substituting the expressions for $a, b, c$ into the perimeter formula: $$ P = 2R\sin A + 2R\sin B + 2R\sin C = 2R(\sin A + \sin B + \sin C) $$ **Step 3:** Apply the given condition. The problem states that the perimeter is 6 times the arithmetic mean of the sines of its angles: $$ P = 6 \times \frac{\sin A + \sin B + \sin C}{3} $$ $$ P = 2(\sin A + \sin B + \sin C) $$ **Step 4:** Determine the circumradius $R$. Equating the two expressions for $P$ from Step 2 and Step 3: $$ 2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C) $$ Since $A, B, C$ are angles of a triangle, $\sin A, \sin B, \sin C$ are all positive, so their sum $\sin A + \sin B + \sin C \neq 0$. We can divide both sides by this sum: $$ 2R = 2 $$ $$ R = 1 $$ **Step 5:** Calculate the angle $A$. Given that side $a = 1$. Using the Sine Rule $a = 2R\sin A$: $$ 1 = 2(1)\sin A $$ $$ \sin A = \frac{1}{2} $$ **Step 6:** Identify possible values for angle $A$. For a triangle, $0^\circ < A < 180^\circ$. The values of $A$ for which $\sin A = \frac{1}{2}$ are: $$ A = 30^\circ \quad \text{or} \quad A = 150^\circ $$