a/sinA=b/sin2A

a/sinA=b/2sinAcosA

a=b/2cosA

cosA=b/2a

=2x6^ 1/2/(2x3)=6^ 1/2/3

A: COSA = 6 1/2/3.

(2)cosA=(b^2+c^2-a^2)/2bc

6^ 1/2x2x2x6^ 1/2c=3x( 15+c^2)

24c=3( 15+c^2)

8c= 15+c^2

c^2-8c+ 15=0

(c-3)(c-5)=0

C-3 = 0 or c-5=0.

c = 3orc = 5

Test: 1.c=3，a+c = 3+3 = 6 & gt； 2x6^ 1/2

/a-c/=/3-3/=/0/= 0 & lt； 2x6^ 1/2

Therefore, the condition of triangle is satisfied,

c=3

2.c=5，a+c = 3+5 = 8 & gt； 2x6^ 1/2

/a-c/=/3-5/=/-2/= 2 & lt； 2x6^ 1/2

Satisfy the conditions of triangle establishment

So c=5

A: c=3orc=5 Two solutions.