Find all missing angles and sides of the triangle with b 2

Find all missing angles and sides of the triangle with b = 2, c = 3, and B = 40degree. If there is more than one solution, find all solutions. If there is no solution, state \"no solution\".

Solution

Given, b=2, c=3 and B=40^o

From Law of Sines, we have

(a/Sin A) = (b/Sin B) = (c/Sin C)

Hence, (b/Sin B) = (c/Sin C) implies that

2/sin 40^o = 3/sin C

Sin C = 1.5*0.6428 = 0.9642

C = 74.62^o

In a triangle, sum of all the angles = 180^o implies that A+B+C = 180^o

Hence A+40+74.62 = 180^o

A = 65.38^o

Hence, (b/Sin B) = (a/Sin A) implies that

(a/sin 65.38) = (2/sin 40)

a = (0.9091*2 / 0.6428) = 2.828

a = 2.83

Hence, a = 2.83, C = 74.62^o & A = 65.38^o