sinA=3/5===>cosA=4/5,cosB=5/13===>sinB=12/13
cosC=cos(180º-(A+B))=-cos(A+B)=sinAsinB-cosAcosB
=(3/5)(12/13)-(4/5)(5/13)=(36-20)/(5*13)=16/65
三角形ABC中,A正弦3/5,B余弦5/13,求C余弦.
三角形ABC中,A正弦3/5,B余弦5/13,求C余弦.
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数学人气:631 ℃时间:2020-02-04 13:34:21
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