Me solve the triangle. a = 32°, a = 19, b = 12 by using the law of sines

Δ ABC: ∠A = 32°  ;  a = 19 (side a opposed to angle A)
and side b = 12 (opposed to Angle B)

Law of sine's
:
(sin A)/19 = (sin B)/12, but sin A = sin 32° = 0.53
0.53/19 = (sin B)/12

and sin B = (0.53 x 12) / 19 → sin B = 0.335. 
Calculate ∠ B: 
sin B = 0.335. → sin⁻¹(0.335) = 19.57° ≈ 20°
∠ A = 32°, ∠B ≈ 20° and ∠C = 180° -32°-20° = 128°
Now that we have got the 3 Angles, let's find c, the 3rd side of the Δ

sin 32/19 = sin 128/c
 0.03 = 0.788/c   and c ≈ 26
+a = 19, b = 12 an c = 26