Question

在三角形ABC中,tan(A+B)/2=2sinC 1.求角C的大小 2.若AB=1,求三角形ABC周长的取值范围

Answer 1

tan(A+B)/2=sinC

tan(180-C)/2=sinC

tan(90-(C/2))=sinC

cot(C/2)=sinC

cotc/2=cos(c/2)/sin(c/2)

sinc=2sin(c/2)cos(c/2)

所以有cos(c/2)/sin(c/2)=2sin(c/2)cos(c/2)

1＝2sin²(c/2)

sin²(c/2)=1/2

sin(c/2)=（根号2）/2

C/2=45

C=90

AB=1,那么AC=AB*cosA,BC=ABsinA

那么周长L=AB+AC+BC=1+sinA+cosA=1+根号2 sin(A+45)

由于0