[" In a "/_ABC,/_A=(2 pi)/(3),b-c=3sqrt(3)" and "Delta=(9sqrt(3))/(2)cm^(2)],[" then "a=(" in "cm)]

In a Delta ABC,A=(2 pi)/(3),b-c=3sqrt(3)cm and area(Delta ABC)=(9sqrt(3))/(2)cm^(2). Then a is

In DeltaABC, A=(2pi)/(3),b-c=3 sqrt(3) cm and area of DeltaABC=(9 sqrt(3))/(2)cm^(2) . Then BC=

In a A B C ,A=(2pi)/3,b-c=3sqrt(3)c m and area ( A B C)=(9sqrt(3))/2c m^2dot then ' a ' is

In DeltaABC, A = (2pi)/(3), b -c = 3 sqrt3 cm and " area of " Delta ABC = (9sqrt3)/(2) cm^(2) , then a =

In DeltaABC, A=(2pi)/(3), b-c=3sqrt3 cm and are (DeltaABC) =(9sqrt3)/(2) cm^(2). Solve for side a.

In DeltaABC, A = (2pi)/(3), b -c = 3 sqrt3 cm and " area of " Delta ABC = (9sqrt3)/(2) cm^(2) , then BC =

In DeltaABC, A = (2pi)/(3), b -c = 3 sqrt3 cm and " area of " Delta ABC = (9sqrt3)/(2) cm^(2) , then BC =

/_A=(pi)/(6) and b:c=2:sqrt(3), find /_B

In a Delta ABC,A=(pi)/(3) and b:c=2:3. If tan alpha=(sqrt(3))/(5),0

If in Delta ABC, b-c = 3sqrt3 , A= 120 ^(@) and the area of Delta ABC = (9sqrt3)/( 2) sq. cm, then a=