Let a, B and C be opposite to each other. If a = π 3 and a = 3, then the value range of B2 + C2 is () A. [3，6]B. [2，8]C. （2，6）D. （3，6]

Let a, B and C be opposite to each other. If a = π 3 and a = 3, then the value range of B2 + C2 is () A. [3，6]B. [2，8]C. （2，6）D. （3，6]

∵ a = π 3, a = 3, from the cosine theorem, we can get 3 = B2 + c2-2bc · cos π 3, ∵ 3 = B2 + C2 BC, ∵ B2 + C2 ≥ 2BC, ∵ BC ≤ B2 + C22 ∵ 3 = B2 + C2 BC ≥ B2 + c2-b2 + C22, the solution is B2 + C2 ≤ 6, if and only if B = C, we take the equal sign, and from 3 = B2 + C2 BC, we can get B2 + C2 = 3 + BC > 3, so the value range of B2 + C2 is: (3, 6] so choose: D

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