Question

已知三角形中，cosA=3/5，cosB=12/13求cosC

Answer 1

因为cosA=3/5 sinA=√(1-cos²A)=4/5
cosB=12/13 sinB=√(1-cos²B)=5/13
所以cosC=cos(π - (A+B))
=-cos(A+B)
=-(cosAcosB-sinAsinB)
=sinAsinB-cosAcosB
=4*5/13*5-3*12/13*5
=-16/65 敬请采纳哦

Answer 2

cosC=cos(180-A-B)=-cos(A+B)
=sinAsinB-cosAcosB
因cosA=3/5，cosB=12/13，所以:sinA=4/5 sinB=5/13
所以:cosC=4/5*5/13-3/5*12/13
=4/13-36/65
=-16/65

Answer 3

sinA=√(1-cos²A)=4/5, sinB=√(1-cos²B)=5/13
cosC=cos(180°-A-B)=-cos(A+B)=-cosAcosB+sinAsinB=-(3/5)*(12/13)+(4/5)*(5/13)=-16/65

Answer 4

cosA=3/5 sinA=√(1-cos²A)=4/5
cosB=12/13 sinB=√(1-cos²B)=5/13

cosC=cos(π - (A+B))
=-cos(A+B)
=-(cosAcosB-sinAsinB)
=sinAsinB-cosAcosB
=4*5/13*5-3*12/13*5
=-16/65

Answer 5

三角函数貌似都是针对直角三角形的吧？貌似cos90=0
莫非C 不是直角？