How do you solve a triangle where side a=30, side b=20 and angle A=50?

We can use the Law of Sines to find angle B

sinB/ 20 = sin50 / 30

sin^-1(20sin(50) / 30) = B = about 30.7°

And angle C = 180 - 50 - 30.7 = 99.3

And side c can be found by using the Law of Sines, again

c sin 99.3 = 30 / sin 50

c = 30 sin 99.3 / sin 50 = about 38.65

Since we have an SSA situation, we can see if we have another possible truangle, thusly :

Subtract angle B from 180 = 180 - 30.7  = 149.3

And when we add this to the known angle of 50 we have more than 180°, so no second triangle exists.