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∵a^2<b^2+c^2,b^2+c^2-a^2>0又∵0<c<b<a∴由余弦定理:cosA=(b^2+c^2-a^2)/2bc>0,△ABC中∠A最大,且0<A<π,所以首先確定0<A<π/2 ∵a>b>c∴A>B>C,2A>B+C=π-A,3A>π,A>π/3 ∴π/3<A<π/2
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