In the diagram ΔABC, ∠ACB = 90°, CD⊥AB, ∠ACD = 30°, AC = 8 cm.
Suppose BD = x and CD = y.
If angle ∠ACD = 30°, then angle ∠BCD = 60°
and angle ∠CAD = 60°, angle ∠CBD = 30°
Now we have two small right triangle ΔCDA and ΔCDB.
In Right Triangle ΔCDA; AC = 8 cm and ∠ACD = 30°.
[tex]cos(30^o) = \frac{CD}{AC} \\CD = AC*cos(30^o)\\y = 8*\frac{\sqrt{3}} {2} \\y = 4\sqrt{3}[/tex]
In Right Triangle ΔCDB; angle ∠BCD = 60° and CD = y = 4√3.
[tex]tan(60^o) = \frac{BD}{CD} \\BD = CD*tan(60^o) \\x = y* \sqrt{3} \\x = 4\sqrt{3} *\sqrt{3} \\x = 4*3\\x=12 \;cm[/tex]
Hence, BD = 12 cm.
