so, two solutions, two triangles. Draw the diagram. CB = CB' = 6 cm = a
we will use two triangles ABC and AB'C
use sine law in triangle AB'Cto find angle B'
B'= 49.88 = 50 degrees.
In triangle CBB', angle B = angle B' = 50 degrees.
So, supplementary angle B = 180 - 50 = 130 degrees.
In triangle ACB, angle C = 180 - (35 + 130) = 15 degrees.
In triangle ACB', angle C = 180 - (35 + 50) = 95 degrees.
Use sin law again in triangle ACB,
AB = 2.7 cm
Use sin law again in triangle ACB',
AB = 10.4 cm
Solution is :
For triangle ABC,
angle A= = 35 degrees, B = 130 degrees, C = 15 degrees
a= 6 cm, b = 8 cm, c = AB = 2.7 cm
For triangle AB'C,
angle A= = 35 degrees, B' = 50 degrees, C = 95 degrees
a= 6 cm, b = 8 cm, c = AB' = 10.4 cm
Show some respect for others trying to help by making your statements clear, and avoid potentialy offensive comments like: "A for a, get it?". If someone more experienced with this stuff has trouble understanding what you write perhaps you should consider that the fault may lie with you rather than with them.