In triangle ABC, if (a + B + C) (B + C-A) = 3bC, then angle a is

The solution is to regard B + C as a whole, simplify and use cosine theorem
(a+b+c)(b+c-a)
=(b+c)^2-a^2
=b^2+c^+2bc-a^2=3bc
∴a^2=b^2+c^2-bc
According to the cosine theorem of any triangle, a ^ 2 = B ^ 2 + C ^ 2 - 2 · B · C · cosa
∴cosA=1/2
∵ A is the inner angle of the triangle
∴A=60°

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