sinB=[tex]\frac{AC}{BC}[/tex]
sin60°=[tex]\frac{\sqrt{3} }{2}[/tex]
[tex]\frac{AC}{BC}=\frac{\sqrt{3} }{2}[/tex]
[tex]\frac{6\sqrt{3} }{BC}=\frac{\sqrt{3} }{2}[/tex]
[tex]BC=\frac{2*6\sqrt{3} }{\sqrt{3} } =12cm[/tex]
∡C=30°⇒AB=BC/2=6cm
BD-bisectoare ⇒∡DBA=60/2=30°
In ΔDBA,∡A=90,∡B=30
cos∡ABD=[tex]\frac{AB}{BD}[/tex]
cos30°=[tex]\frac{\sqrt{3} }{2}[/tex]
[tex]\frac{AB}{BD}=\frac{\sqrt{3} }{2}[/tex]
[tex]\frac{6}{BD}=\frac{\sqrt{3} }{2}[/tex]
[tex]BD=\frac{2*6}{\sqrt{3} }=\frac{12}{\sqrt{3} }=4\sqrt{3}[/tex]
AD=BD/2=2√3
