Question

在三角形ABC中，sinA=5/13,cosB=3/5,求cosC的值 提示：此题有两解

Answer 1

1)当A,B都为锐角时,
cosA=√(1-sin²A)=12/13,
cosB=√(1-sin²B)=4/5,
A+B=180-C,
cos(A+B)=cos(180-C)=-cosC,
cosC=-cos(A+B)=-[cosA*cosB-sinA*sinB]=-33/65,

2)当A钝角，B为锐角时，
cosA=-√(1-sin²A)=-12/13,
cosB=√(1-sin²B)=4/5,
cosC=-cos(A+B)=-[cosA*cosB-sinA*sinB]=63/65

所以cosc的值为:-33/65,或63/65.

Answer 2

sinA=5/13 => cosA=12/13或cosA=-12/13
cosB=3/5 => sinB=4/5
cosC=cos(PI-A-B)=-cos(A+B)=sinAsinB-cosAcosB
代入上述值得：
1、当cosA=12/13时
cosC=5/13*4/5-12/13*3/5=-16/65
2、当cosA=-12/13时
cosC=5/13*4/5-(-12/13*3/5)=56/65

Answer 3

sinA=5/13,cosA=±12/13
cosB=3/5;sinB=4/5
cosC=-cos(A+B)=-cosAsinB+cosBsinA=-33/65或63/65