快回复已知△ABC的三内角A,B,C所对边的长分别为a,b,c,设向量p=(a+c,b),q=(b-a,c-a),若p/

p//q----> (a+c)*(c-a) -b(b-a)=0
c^2-a^2-b^2 +ab=0------>a^2+b^2-c^2= ab
cosC = (a^2+b^2-c^2)/2ab= ab/2ab =1/2 ---->C=60度.
s=1/2 ac sinB =1/2 *c* 根号3/2 = 根号3/2
c=2
cosB =( a^2+b^2-c^2)/2ab= 1/2
(1+b^2 -4)/2b =1/2------>b=(1+根号13）/2
c/sinC= b/sinB-----

p//q
=> (a+c)/b = (b-a)/(c-a)
c^2-a^2 = b^2 -ab
c^2=a^2+b^2- ab
By cosine-rule
=> 2abcosC = ab
cosC = 1/2
C =π/3

(1) p//q 得 (a+c)/b=(b-a)/(c-a)
c^2-a^2=b^2-ab
c^2=a^2+b^2-ab=a^2+b^2-2abcosC
得cosC=1/2 C=60
(2) BC边上的高AD ＝h
sABC=1/2*a*h(h是其高)=根号3/2
h=根号3
BD=1
就是D 是C点
C=90
c=AB=2
c/sinC=2/sin90=2