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[几何] 三角形中满足$\frac a{\sin B}+\frac b{\sin A}=2c$,求角$A$

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isee Posted 2021-9-28 22:01 |Read mode
比这题弱:三角形中sinA/cosB+sinB/cosA=2


题:三角形中满足$\frac a{\sin B}+\frac b{\sin A}=2c$,求角$A$.

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 Author| isee Posted 2021-9-28 22:03
Last edited by isee 2021-9-29 11:20我是硬化边计算的,条件去分母等价于
$a^2+b^2=2bc\sin A$,又有 $b^2+c^2-a^2=2bc\cos A$,
于是 $(a^2+b^2)^2+(b^2+c^2-a^2)^2=4b^2c^2$,
整理为$(a^2+b^2-c^2)^2+(a^2-b^2)^2=0$,斜边为$c$的等腰直角三形角,$A=\frac \pi4$.

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kuing Posted 2021-9-29 00:26
化角把逼一夹:
\[2\leqslant\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A}=2\sin C\leqslant2,\]就这么简单

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