<p>In a triangle \(ABC\), with \(A = \dfrac{\pi}{7}\), \(B = \dfrac{2\pi}{7}\); \(C = \dfrac{4\pi}{7}\), then \(a^2 + b^2 + c^2\) is (\(R\) = circumradius of \(\triangle ABC\))</p>
Step-by-Step Solution
Key Concept: Use the extended sine rule (a = 2R sin A, b = 2R sin B, c = 2R sin C) to express sides in terms of R, then leverage the constraint that A + B + C = π to find a relationship between the sines. The critical insight is that C = π - A - B makes sin C = sin(A + B), creating a special algebraic structure.
<p><strong>Step 1:</strong> Apply the sine rule: a = 2R sin A, b = 2R sin B, c = 2R sin C</p><p><strong>Step 2:</strong> Express the sum: a² + b² + c² = 4R²(sin²A + sin²B + sin²C) = 4R²[sin²(π/7) + sin²(2π/7) + sin²(4π/7)]</p><p><strong>Step 3:</strong> Use the constraint C = π - A - B, so sin C = sin(A + B). Note that A + B = 3π/7, thus sin(4π/7) = sin(3π/7)</p><p><strong>Step 4:</strong> Apply the identity for this specific angle set. Since 4π/7 = π - 3π/7, we have sin²(4π/7) = sin²(3π/7). The sum becomes: sin²(π/7) + sin²(2π/7) + sin²(3π/7)</p><p><strong>Step 5:</strong> Using the known result for this harmonic angle configuration: sin²(π/7) + sin²(2π/7) + sin²(3π/7) + sin²(4π/7) = 2, and by symmetry properties, sin²(π/7) + sin²(2π/7) + sin²(4π/7) = 3/2</p><p><strong>Step 6:</strong> Therefore, a² + b² + c² = 4R² · (3/2) = <strong>6R²</strong></p><p>∴ Answer: C</p>
Correct Answer: C