Question

在三角形ABC中,sinA·sinB=sinC的平方-sinA的平方-sinB的平方，求角C

Answer 1

答：
sinA·sinB=(sinC)^2-(sinA)^2-(sinB)^2
=(sinC)^2-(sinA)^2-{sin[π-(A+C)]}^2
=(sinC)^2-(sinA)^2-[sin(A+C)]^2
=(sinC)^2-(sinA)^2-(sinAcosC+cosAsinC)^2
=(sinC)^2-(sinA)^2-[(sinA)^2(cosC)^2+2sinAcosAsinCcosC+(cosA)^2(sinC)^2]
=(sinC)^2-(sinA)^2-(sinA)^2(cosC)^2-2sinAcosAsinCcosC-(cosA)^2(sinC)^2
=(sinC)^2[1-(cosA)^2]-(sinA)^2-(sinA)^2(cosC)^2-2sinAcosAsinCcosC
=(sinC)^2(sinA)^2-(sinA)^2-(sinA)^2(cosC)^2-2sinAcosAsinCcosC
=-(sinA)^2(cosC)^2-(sinA)^2(cosC)^2-2sinAcosAsinCcosC
=-2sinAcosC(sinAcosC+cosAsinC)
=-2sinAcosCsin(A+C)
= -2sinAcosCsinB
所以：
cosC=-1/2
C=120°