Referring to qwr's figure, the SSA triangle ("ambiguous case") can be resolved using an "altitude test". The green segment is the altitude of the triangle(s) $ \ ABC \ $ because it is perpendicular to the unknown base $ \ b \ $ . The trig we've had for right triangles tells us that $ \ h \ = \ c \ \sin A \ , $ for which we have been given values.
We now compare our "loose side" $ \ a \ , $ which can be thought of as if it were on a "hinge" at point $ \ B \ , $ to this altitude:
If $ \ a \ = \ h \ , $ then that side is the altitude of the triangle, making $ \ ABC \ $ a right triangle; we can then use the trigonometry of right triangles to finish solving for the remaining unknown angle $ \ B \ $ and the base $ \ b \ ; $
if $ \ a \ < \ h \ , $ then it is simply too short to reach the "base" of this figure, and thus cannot form a closed triangle: in this case, there is no triangle to be solved;
if $ \ h \ < \ a \ < \ c \ , $ then the side $ \ a \ $ is more than long enough to reach the base and form a closed triangle; however, there are two possible places where it could contact the base (at $ \ C \ $ and $ \ C' \ $ ) in qwr's diagram; we know $ \ a \ $ and $ \ \sin A \ , $ so we can use the Law of Sines to work out angle $ \ C \ $ , since we know the length of $ \ c \ $ ; the trouble is that there are two possible angles $ \ C \ $ that both have the same value of $ \ \sin C \ , $ one in the first quadrant, the other in the second; in this case, both solutions are valid and we have two possible triangles, one with side $ \ a \ $ extended away from side $ \ c \ $ (making $ \ C \ $ an acute angle), the other where it is "tucked under" side $ \ c \ $ (for which $ \ C \ $ is an obtuse angle) ;
finally, if $ \ c \ < \ a \ , $ then side $ \ a \ $ is again quite long enough to reach the base, but too long to "tuck under" side $ \ c \ $ ; in this case, we can only have the solution for angle $ \ C \ $ where it is acute (side $ \ a \ $ being "extended" away from side $ \ c \ $ ) .
Believe me, this is the hardest portion of learning "solutions of triangles" (having taught trig a number of times); by comparison, all the other triangle types are fairly simple to work with.
For the problem you show, $ \ h \ = \ 100 \ \sin 26º \ \approx \ 43.8 \ . $ Side $ \ a \ $ is longer than that, but shorter than side $ \ b \ , $ so this is a "two-triangle" case, with two values for angle $ \ B \ $ . I take it you can proceed from there...