Question

三角形中cosAsinB+cosAcosB+cosC=0,求A

cosA(sinB+cosB)+cosC=0,求A

Answer 1

cosAsinB+cosAcosB+cosC

=cosAsinB+cosAcosB+cos[π-(A+B)]

=cosAsinB+cosAcosB-cos(A+B)

=cosAsinB+cosAcosB-cosAcosB+sinAsinB

=cosAsinB+sinAsinB

=sinB(sinA+cosA)

=√2sinBsin(A+π/4)=0

所以可得：sinB=0(不成立)或者sin(A+π/4)=0

所以：A+π/4=0或A+π/4=π

所以：A=-π/4或A=3π/4

因为A为三角形内角，A=-π/4舍去，所以A=3π/4

Answer 2

3派/4 cosAsinB+cosAcosB=-cosC=cosAcosB-sinAsinB, 得cosA=-sinA