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根据ABC中的等式tgA/2tgB/2+tgB/2tgC/2+tgC/2tgA/2=1,可得到一个重要的不等式:在△ABC中,tgA/2+tgB/2=tgC/2≥3~(1/2).
According to the equation tgA/2tgB/2+tgB/2tgC/2+tgC/2tgA/2=1 in ABC, an important inequality is obtained: in ΔABC, tgA/2+tgB/2=tgC/2≥ 3~(1/2).