Suppose in triangle ABC, you know side a = 10, b = 16 and angle A is 30 degrees
By the law of sines,
SinB/16 = Sin30/10
10*sinB = (1/2)!6
SinB = 0.8, therefore B = 53.1 degrees. But sine is also positive in the second quadrant, so there is a possible second triangle with B = 126.9 degrees. This can work because C would equal 23.1 degrees in this triangle and C = 96.9 in the first triangle
If angle A was 60 degrees and B came out to 53.1 or 126.9 then only 1 triangle exists since (126.9 + 60 = 186.9) angle C + angle A > 180.
If you try to solve for an angle an get Sin > 1 or < -1 then there are no solutions.