在△ABC中,内角A,B,C所对的边分别是a,b,c,已知tanA=sinC/2-cosC,c=3.(1)求b/a; (2)若△ABC的面积为3,求cosC.
在△ABC中,内角A,B,C所对的边分别是a,b,c,已知tanA=
,c=3.
(1)求
;
(2)若△ABC的面积为3,求cosC.
sinC |
2-cosC |
(1)求
b |
a |
(2)若△ABC的面积为3,求cosC.
数学人气:267 ℃时间:2019-10-18 08:18:30
优质解答
(1)tanA=sinAcosA=sinC2-cosC,即2sinA-sinAcosC=cosAsinC,整理得:2sinA=sinAcosC+cosAsinC=sin(A+C)=sinB,利用正弦定理asinA=bsinB化简得:2a=b,则ba=2;(2)∵2a=b,△ABC面积为3,c=3,∴S△ABC=12absin...
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