设三角形ABC的内角A,B,C的对边长分别为a,b,c,且A=60°,c=3b.求a/c的值.求tanB+tanC的值

a²=b²+c²-2bccosA
a²=c²/9+c²-2c²/3*(1/2)
a²=7c²/9
a²/c²=7/9
a/c=√7/3
设b=t,c=3t,a=√7t
cosB=(a²+c²-b²)/2ac
=15/(6√7)
=5√7/14
tanB=√3/5
cosC=(a²+b²-c²)/2ab
=-1/(2√7)
=-√7/14
tanC=-3√3
tanB+tanC=√3/5-3√3=-14√3/5