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In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and satisfy 2bcosa = ccosa + acosc. (1) find the size of angle a; (2) if a = 3 and s △ ABC = 334, try to judge the shape of △ ABC and explain the reason
(1) From 2bcosa = ccosa + acosc and sine theorem, it is obtained that 2sinbcosa = sin (a + C) = SINB, that is SINB (2cosa-1) = 0, ∵ 0 < B < π, ∵ SINB ≠ 0, ∵ cosa = 12, ∵ 0 < a < π, ∵ a = π 3; (2) ∵ s △ ABC = 12bcsina = 334, that is 12bcsin π 3 = 334, ∵ BC = 3, ①∵ A2 = B2 + c2-2bccosa, a = 3, a = π 3, ∵ B2 + C2 = 6, ② B = C = 3, then △ ABC is an equilateral triangle
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