Answer:
[tex]AB=6\sqrt{3}[/tex]
Step-by-step explanation:
[tex]Given\ \angle A=30\textdegree\\\\AC=12\\\\\cos \theta=\frac{adjacent}{hypotenuse}\\\\\cos A=\frac{AB}{AC}\\\\\cos 30=\frac{AB}{12}\\\\\frac{\sqrt{3}}{2}=\frac{AB}{12}\\\\AB=6\sqrt{3}[/tex]
