IMO D

my way of solving is not a GMAT way
drop a perpendicular from B to AD say at o

statement1)
BD=6rut3

let oc=x
in tri bod
cos30=od/bd
cos 30=rut3/2
solving x=3
applying pythagoras theorem in tri BOD
we ob as 3rut3

applying same in tri boc
bc=6

suff

statement 2)
in tri boc
tan 60=ob/x
hence ob=xrut3........1
in tri BOD
tan 30=ob/(x+6)
solving ob=(x+6)/rut3.........2

solving eqn 1 and 2 x=3

rest aplying pythagoras we get BC
suff

hence D

It should be D

Stmt 1:
Draw a perpendicular from B to AD and mark that point as O then triangle BOD is 30-60-90 right triangle
we know one of the side BD so we can find out other two sides as side of the 30-60-90 right trinangle are 1:sqr3:2
so we know BO and OD
OD - 6 will give us OC
applying Pythagoras on triangle BOC we can find BC

Sufficient

Stmt 2: with BCA = 60 and CDB = 3= CBD should be 30 and sides cd should be equal to bc (sides opp equal angles)

Sufficient

Ans D