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 a =

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

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

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

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

If, in DeltaABC, a = sqrt(2), b = sqrt(3), c = sqrt(5) , then the area of Delta ABC is

In Delta ABC angle A = (pi)/(6) & b:c= 2: sqrt(3) "then " angle B is

In A Delta ABC if ( sqrt(3)-1)a=2b, A=3B then c is

If Delta ABC ~ Delta DEF, BC = 3cm, EF = 4cm, and Area of Delta ABC = 54 cm^(2) , then Area of Delta DEF is