Answer:
[tex]\sqrt{3}[/tex]
Step-by-step explanation:
In ΔABC ∠C=90° , ∠A =30°
Let ∠B =[tex]\alpha[/tex]
∠A+∠B+∠C = 180°
⇒90° + 30° +[tex]\alpha[/tex]° =180°
⇒[tex]\alpha[/tex] =60°
Now, in ΔACD [tex]sin(30)=\frac{CD}{6}[/tex] [tex]=0.5[/tex]
CD =3
In ΔCDB [tex]tan(60)=\frac{3}{BD}[/tex]=[tex]\sqrt{3}[/tex]
BD =[tex]\sqrt{3}[/tex]
