In the triangle ABC, if a ^ 2 + 3bC = (B + C) ^ 2, then the angle A=

From cosine theorem
a^2=b^2+c^2-2bccosA
From the question
a^2+3bc=b^2+c^2+2bc
Simplify
a^2=b^2+c^2-bc
So cosa = - 1 / 2
And because in the a triangle
So a = 120 degrees

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