在三角形ABC中,内角A,B,C的对边分别是a,b,c，a等于2倍根号三，tan2分之a+b加tan2分之c等于4

Question

2sinBcosC=sinA，求A，B及b，c

Answer 1

∴ tan(A+B)/2=cotC/2
∴cotC/2+tanC/2=4.
(cosC/2/sinC/2)+sinC/2/cosC/2=4.
[cos^2(C/2)+sin^2(C/2)]/sin(C/2)cos(C/2)=4.
1/(2sin(C/2)cos(C/2)=2.
sinC=1/2
2sinBcosC=sinA.
=sin(B+C).
=sinBcosC+cosBsinC.
sinBcosC-cosBsinC=0.
sin(B-C)=0.
B-C=0.
B=C.
A+B+C=180°.
A+2B=180°.
2B=180°-30°.
=150°.
B=C=75°
应用正弦定理：a/sinA=c/sinC.
c=asinC/sinA.
=2√3*sin75°/sin30°.
=2√3(√2/4)(√3+1)/(1/2).
=3√2+√6.
∴A=30.
B=75,
C=75,
c=3√2+√6.