在三角形abc中已知b²;-bc-2c²;=0且a=根号6,cosA=8分之7,则三角形ABC的面积为?
在三角形abc中已知b²;-bc-2c²;=0且a=根号6,cosA=8分之7,则三角形ABC的面积为?
数学人气:685 ℃时间:2019-08-20 05:27:49
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余弦定理:a^2=b^2+c^2-2bccosA6=b^2+c^2-7bc/4已知b²-bc-2c²=0消去b^2:c^2=2+bc/4代入:b²-bc-2c²=0b^2=4+3bc/2(bc)^2=(2+bc/4)*(4+3bc/2)5(bc)^2-32bc-64=0bc=8三角形ABC的面积S=sinA*bc/2=(1...
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