To get a better understanding, let's sketch the geometry of the 
situation. I start by drawing an arbitrarily long line labeling 
the endpoint as the vertex of the given angle, in this case B.



At point B we draw the angle B = 10° by drawing side 
BC = a = 19:



Next we take a compass and open it to the length
b = CA = 10.  Put the sharp point on C and swing 
an arc that cuts the horizontal line that we drew
first in two places:



So it appears in this case there are two choices for point A.  
Either this way with the green line:


 
or this way with the blue line:



Now let's do the calculating using the law of sines:



substitute a=19, b=10, B=10°



We use only the first and second parts:



Cross-multiply



Divide both sides by 10



Do the calculation:



[If this had come out greater than 1 there would have 
been 0 solutions.  If this had come out equal to 1 
there would have been 1 solution, and it would have 
been a right triangle with hypotenuse b.]

Since it came out less than 1, we can tell there is
either 1 or 2 solutions.

First solution:
Next we calculate the first quadrant (acute) solution
for angle A using inverse sine feature on the calculator
and get angle A = 19.26462015°.  Then we can calculate the
third angle C by adding 19.26462015° and 10° and subtracting
from 180°, getting angle C = 150.7353799°.

Now we use the law of sines again:



This time we use only the second and third parts,
and substitute 150.7353799° for angle C 



Cross-multiply:



Divide both sides by sin(10°)





-----------------

Next we see if there is a second solution.
We calculate the second quadrant (obtuse) solution
for angle A by finding the complement of 19.26462015°,
by subtracting it from 180°, getting 160.7353798°.
Then we can calculate the third angle C by adding 
160.7353798° and 10° and subtracting
from 180°, getting angle C = 9.2646202°.

[If this had come out negative there would have 
been only the 1 solution.]

So we have a second solution, and we use the law of
sines to find c just as we did to find the first
solution:



Cross-multiply:



Divide both sides by sin(10°)





Edwin