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第一問,如下:(b^2+c^2-a^2)tanA=√3bc, 由余弦定理b^2+c^2-a^2=2bccosA代入,有:2bcsinA=√3bc, 即:sinA=√3/2可知:A=60° 題目的第二問,是不是有問題啊?A不是求出來了麼?
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