In triangle $ABC$,if $C=120^{\circ}$,$c=\sqrt{19}$,and $b=3$,then $a=$

The smallest angle of the triangle whose sides are $6 + \sqrt{12}$,$\sqrt{48}$,and $\sqrt{24}$ is:

The sides of a triangle are $\sin \alpha$,$\cos \alpha$,and $\sqrt{1 + \sin \alpha \cos \alpha}$ for some $0 < \alpha < \frac{\pi}{2}$. Then the greatest angle of the triangle is.....$^o$

In a $\Delta ABC$,if $A = 30^o$,$b = 2$,and $c = \sqrt{3} + 1$,then $\frac{C - B}{2} = $ ....$^o$

In a triangle $ABC$,if $a-2b+c=0$,then $\cot \left(\frac{A}{2}\right) \cdot \cot \left(\frac{C}{2}\right) = $

In a $\Delta ABC$,$a = 5, b = 4$ and $\cos(A - B) = \frac{31}{32}$,then side $c$ is equal to