here you see, a<b, so calculate,
,
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