Trignometry - Cosine Formulae c = 21
b = 19
B= $65^o$
solve A with cosine formulae
$a^2+21^2-19^2=2a(21)cos65^o$
yield an simple quadratic equation in variable a
but, $\Delta=(-2(21)cos65^o)^2-4(21^2-19^2) < 0$ implies the triangle as no solution?
How to make sense of that? Why does this happen and in what situation? Please give a range, if any, thanks.

Here is an investigation without directly using sine law:
From A draw a perpendicular line to BC. Let the intersection be H
BH is AB*cos(65) which is nearly 8.87
From Pythagoras law, AH is nearly 19.032
AC is 19 But we arrived at a contradiction since hypotenuse is less than another side which then immediately implies $\sin(C)>1$    
