This is a problem of the AMBIGUOUS case SSA, (side-side-angle) 
where you are asked to solve a triangle given two sides and a 
non-included angle.  If I have drawn this triangle accurately
to scale, then it appears that there are two solutions to
this problem.  If we open a compass to 19 units, put the sharp
point of the compass at C and swing an arc, like this,



So it appears that there are two solutions, these two triangles:



We will use the law of sines to determine angle A



b·sin(A) = a·sin(B)

sin(A) = 

sin(A) = 

sin(A) = 0.948918408  <--- If this had been greater than 1, there
                           would have been no solution.  If this had 
                           been exactly 1, there would have been 1 right
                           triangle angle solution.  But since it is less 
                           than 1, we can tell that there is either 1 or 2
                           solutions.  
                
If you use your calculator with the inverse sine,
you get 

A = 71.5303934°.

That is a correct value for angle A.  However there is another 
possible angle with that same sine, which is a second quadrant 
angle, and it is found by subtracting A = 71.5303934° from 180°.
We will use primes on the letters to indicate the possible
second solution: 

A' = 180° - 71.5303934° = 108.4696066°

So we put those two values in:

First solution               Second solution (maybe)

A = 71.5303934°              A' = 108.4696066°
B = 55°                      B' = 55°  
C =                          C' = 
a = 22                       a' = 22  
b = 19                       b' = 19  
c =                          c' =  

To find out whether there are 2 solutions or only 1,
we calculate the angles C and C', and see if both
are possibilities:

We calculate C by using 180°-A-B = 180°-71.5303934°-55° = 53.4696066°

We calculate C' by using 180°-A'-B' = 180°-108.4696066°-55° = 16.5303934°

Since C' came out positive, we know that there are two solutions,
[If C' had come out negative, there would have been only 1 solution].

First solution               Second solution

A = 71.5303934°              A' = 108.4696066°
B = 55°                      B' = 55°  
C = 53.4696066°              C' = 16.5303934° 
a = 22                       a' = 22  
b = 19                       b' = 19  
c =                          c' = 

Now we just have to calculate sides c and c'

To calculate side c:

 = 

 =  

c·sin(55°) = 19·sin(53.4696066°)  

c = 

c = 18.63791123

---

To calculate side c':

 = 

 =  

c'·sin(55°) = 19·sin(16.5303934°)  

c' = 

c' = 6.599451972

So we end up with:

First solution               Second solution 

A = 71.5303934°              A' = 108.4696066°
B = 55°                      B' = 55°  
C = 53.4696066°              C' = 16.5303934° 
a = 22                       a' = 22  
b = 19                       b' = 19  
c = 18.63791123              c' = 6.599451972

Incidentally the drawings above are to scale,
because I calculated them before drawing them.

You can round the solutions to tenths. 

A = 71.5°                    A' = 108.5°
B = 55°                      B' = 55°  
C = 53.5°                    C' = 16.5° 
a = 22                       a' = 22  
b = 19                       b' = 19  
c = 18.6                     c' = 6.6


However, you should wait until the end to round off
because if you round earlier, the answers will likely
be too inaccurate. 

Edwin