Question

在三角型ABC中,A.B.C对应的边为a.b.c,a=2倍根号3,tan[(A+B)/2]+tan(c/2)=4,2si

Answer 1

tan[(A+B)/2]+tan(C/2)=4

tan[(π-C)/2]+tan(C/2)=4

cot(C/2)+tan(C/2)=4

cos(C/2)/sin(C/2) + sin(C/2)/cos(C/2)=4

1/sin(C/2)cos(C/2)=4

sin(C/2)cos(C/2)=1/4

sinC=1/2

C=30°或150°

2sinBcosC=sinA

2bcosC=a

2abcosC=a^2

a^2+b^2-c^2=a^2

b=c

所以B=C

所以C=30° B=30°

A=120°

正弦定理得

a/sinA=b/sinB

b=asinB/sinA

=2

c=2