在△ABC中a²+b²-mc²=0(m为常数)且cosA/sinA+cosB/sinB=cosC/sinC,求m的值 求回答
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![在△ABC中a²+b²-mc²=0(m为常数)且cosA/sinA+cosB/sinB=cosC/sinC,求m的值 求回答](/uploads/image/z/3008434-58-4.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%ADa%26%23178%3B%2Bb%26%23178%3B-mc%26%23178%3B%3D0%28m%E4%B8%BA%E5%B8%B8%E6%95%B0%29%E4%B8%94cosA%2FsinA%2BcosB%2FsinB%3DcosC%2FsinC%2C%E6%B1%82m%E7%9A%84%E5%80%BC+%E6%B1%82%E5%9B%9E%E7%AD%94)
在△ABC中a²+b²-mc²=0(m为常数)且cosA/sinA+cosB/sinB=cosC/sinC,求m的值 求回答
在△ABC中a²+b²-mc²=0(m为常数)且cosA/sinA+cosB/sinB=cosC/sinC,求m的值 求回答
在△ABC中a²+b²-mc²=0(m为常数)且cosA/sinA+cosB/sinB=cosC/sinC,求m的值 求回答
你好这题简答
如下由cosA/sinA+cosB/sinB=cosC/sinC
得[(b²+c²-a²)/2bc]/a+[(a²+c²-b²)/2ac]/b=[(b²+a²-c²)/2ab]/c
即(b²+c²-a²)/2bca+(a²+c²-b²)/2acb=[(b²+a²-c²)/2abc
即b²+c²-a²+a²+c²-b²=b²+a²-c²
即3c²=a²+b²
即a²+b²-3c²=0
又有a²+b²-mc²=0
即m=3