若△ABC的面积为S=(a²+b²-c²):4根号3,则角C=?
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![若△ABC的面积为S=(a²+b²-c²):4根号3,则角C=?](/uploads/image/z/1603726-70-6.jpg?t=%E8%8B%A5%E2%96%B3ABC%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BAS%3D%EF%BC%88a%26%23178%3B%2Bb%26%23178%3B-c%26%23178%3B%EF%BC%89%EF%BC%9A4%E6%A0%B9%E5%8F%B73%2C%E5%88%99%E8%A7%92C%3D%3F)
若△ABC的面积为S=(a²+b²-c²):4根号3,则角C=?
若△ABC的面积为S=(a²+b²-c²):4根号3,则角C=?
若△ABC的面积为S=(a²+b²-c²):4根号3,则角C=?
∵S=(a²+b²-c²):4√3
又S=1/2absinC
∴1/2absinC=(a²+b²-c²):4√3
∴2√3absinC=a²+b²-c²
∴√3sinC=(a²+b²-c²)/(2ab)
根据余弦定理
cosC=(a²+b²-c²)/(2ab)
∴√3sinC=cosC
∴tanC=sinC/cosC=√3/3
∴C=30º