Question

三角形面积题目

在三角形ABC中，A,B,C所对的角为a,b,c
当三角形面积S=c^2-(a-b)^2,则tanC=?
（选项 ）A1/2 B1/4 C1/8 D1

Answer 1

s=c^2-(a-b)^2=c^2-(a^2+b^2)+2ab
s=1/2absinC
c^2-(a-b)^2=c^2-(a^2+b^2)+2ab=1/2absinC 即 c^2=a^2+b^2-2ab-1/2absinC
又c^2=a^2+b^2-2*a*b*CosC (余弦定理)
所以
4cosC=sinc-4
又 sinC^2+cosC^2=1
求出cosC=-1（舍）或-15/17
sinC=8/17
tanC=-8/15

没选项啊 哥哥 过程肯定是没错

Answer 2

S=1/2 (absinC)，sinC=2S/ab=2[c²-(a-b)²]/ab；
c²=a²+b²-2abcosC，cosC=(a²+b²-c²)/2ab；
tanC=sinC/cosC={2[c²-(a-b)²]/ab}/[(a²+b²-c²)/2ab]=4(c²-a²-b²+2ab)/(a²+b²-c²)
=-4+[8ab/(a²+b²-c²)]

Answer 3

c^2=a^2+b^2-2abcosc
s=c^2-(a-b)^2,=2ab（1-cosc）=1//2absinc
16(1-2cosc+cosc方)=1-cosc方
cosc=1或cosc=15/17