在△ABC中(b+a)/a=SinB/(SinB-SinA)三角形形状计算
在△ABC中(b+a)/a=SinB/(SinB-SinA)三角形形状计算
在△ABC中(b+a)/a=SinB/(SinB-SinA)且Cos2C+CosC=1-Cos(A-B),试判断△ABC的形状,
在△ABC中(b+a)/a=SinB/(SinB-SinA)且Cos2C+CosC=1-Cos(A-B),试判断△ABC的形状,
数学人气:459 ℃时间:2020-01-30 14:08:18
优质解答
(b+a)/a=sinB/(sinB-sinA)(sinB+sinA)/sinA = sinB/(sinB-sinA)(sinB+sinA)(sinB-sinA) = sinAsinB .(1)cos2C+cosC = 1-cos(A-B)1-2sin^2C + cos(180°-A-B) = 1-cos(A-B)1-2sin^2C - cos(A+B) = 1-cos(A-B)cos(A-B)...
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