It is known that the three sides of △ ABC are a, B, C respectively, and | B + C-2A | + (B + C-5) 2 = 0. The value range of B is obtained

It is known that the three sides of △ ABC are a, B, C respectively, and | B + C-2A | + (B + C-5) 2 = 0. The value range of B is obtained

B + C-2A = 0, B + C-5 = 0, solution: B + C = 5, substituting B + C = 5 into B + C-2A = 0, solution: 5-2a = 0, solution: a = 2.5, then C = 5-b, according to the trilateral relationship of triangle: | 5-b-2.5 | B and B < 5-b + 2.5, that is, 2.5-b < B < 2.5 + 5-b, solution: 54 < B < 154