Question

已知三角形ABC的三边a,b,c和面积S满足S=c^2-(a-b)^2,则1-cosC/sinC的值为

Answer 1

由余弦定理得

cosC=(a²+b²-c²)/(2ab)

a²+b²-c²=2abcosC

S=c²-(a-b)²

=c²-a²-b²+2ab

=-(a²+b²-c²)+2ab

=-2abcosC+2ab

=2ab(1-cosC)

又由三角形面积公式得S=(1/2)absinC,因此

(1/2)absinC=2ab(1-cosC)

2(1-cosC)=(1/2)sinC

(1-cosC)/sinC=1/4