This question is about the "ambiguous case" when you are given the lengths of two sides of a triangle and one angle, NOT the angle between the two sides. Constructively you can do this. Draw a line and mark it to the length of one of the given sides. Construct the given angle at one end of that side. At the other end, set compasses to the length of the second side and mark off an arc at that distance.
There are three possibilities:
1) The arc might not cross the second line at all- the radius is too short. There is NO such triangle.
2) The arc might be tangent to the second line. This is the case where there is exactly one triangle- and it is a right triangle.
3) The arc might cross the second line in two different points. This is the case where there are two such triangles.
Frankly, the simplest way to do determine whether there is no such triangle, one, or two is to actually try to solve the triangle. The standard way to solve this problem is to use the cosine law. The cosine law, an extension of the Pythagorean theorem says, that, in any triangle, with side lengths a, b, c and angle C opposite side c, then c^2= a^2+ b^2- 2ab cos(C). In this situation, we are given the lengths a and c, and the angle, C. Finding the third side, b, is equivalent to solving the quadratic equation b^2- (2a cos(C))b+ (a^2- c^2)= 0.
We can solve that using the quadratic formula:
\frac{2a cos(C)\pm \sqrt{4a^2 cos^2(C)+ 4a^2- 4c^2}}{2}= a cos(C)\pm\sqrt{a^2 cos^2(C)+ a^2- c^2}.
Whether there is zero, one, or two solutions to that, depends on the discriminant a^2 cos^2(C)+ a^2- c^2. If that is positive there must be two solutions and so two such triangles.