Solution:
A is the right tringle ABD, C is bisects AD.
AC = CD and BD = 12
BC=BDsin(BDC)
BC=12Sin(45)
[tex]=12\times \frac{1}{\sqrt{2}}=\frac{12}{\sqrt{2}}[/tex]
[tex]AB=\frac{BC}{sin(bac)}=\frac{\left | 2 \right |\sqrt{2}}{sin30}=\frac{\left | 2 \right |\sqrt{2}}{\frac{1}{2}}=\frac{12\times 2}{\sqrt{2}}[/tex]
[tex]AB=\frac{24}{\sqrt{2}}=12\sqrt{2}[/tex]
