∵钝角三角形中有两个是锐角,∴不妨设A,B是锐角
k=(sinA)^2+(sinB)^2+(sinC)^2
=(1-cos2A)/2+(1-cos2B)/2+1-(cosC)^2=2-(cos2A+cos2C)/2-(cosC)^2
=2-cos(A+B)cos(A-B)-(cosC)^2
=2+cosCcos(A-B)-(cosC)^2
=2+cosC[cos(A-B)+cosC]
=2+cosCcos[(A-B+C)/2]cos[(A-B-C)/2]
∵-90