Question

△ABC中,sinC=(sinA+sinB)/(cosA+cosB),求三角形的形状

Answer 1

sinC=(sinA+sinB)/(cosA+cosB)={2 sin[(A+B)/2] cos[(A-B)/2]}/ {2 cos[(A+B)/2] cos[(A-B)/2]}

=sin[(A+B)/2] /cos[(A+B)/2]=sin(π/2-C/2)/cos(π/2-C/2)=cosC/2/sinC/2,

2sinC/2cosC/2=cosC/2/sinC/2,2sin²C/2=1,cosC=0,C=π/2,三角形为直角△.