△ABC中,A=π/6,（1+根号3）c=2b,求角C,若向量CB乘向量CA=1+根号3 求 边a.b.c

(1)由正弦定理
(1+√3)/2=b/c=sinB/sinC=sin(120°-C)/sinC=(√3/2*cosC+1/2*sinC)/sinC
∴tanC=1
∴C=45°
(2)
∵CB/CA=sinA/sinB=sin60°/sin75°=2√3/(√6+√2)
∵向量CB*向量CA=CB*CA*cosC=1+√3
∴CB*CA=√6+√2
∴CB^2=2√3
即CB=4次√12=a
另外相应地求出b=4次√3+1/(4次√3),c=2/(4次√3)