In the adjoining figure, ABC is right-angled triangle at B and ABD is right angled triangle at A. If BD ⊥ AC and BC = $2\sqrt{3}$ cm, find the length of AD.

In △ABC,

tan 30° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{BC}{AB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{2\sqrt{3}}{AB} \\[1em] \Rightarrow AB = 6 \text{ cm}.$

In △ABE,

90° = 30° + ∠ABE [As, exterior angle is equal to sum of two opposite interior angles]

∠ABE = 90° - 30° = 60°.

From figure,

∠ABD = ∠ABE = 60°

In △ABD,

tan ∠ABD = tan 60° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

$\Rightarrow \sqrt{3} = \dfrac{AD}{AB} \\[1em] \Rightarrow AD = AB\sqrt{3} \\[1em] \Rightarrow AD = 6\sqrt{3} \text{ cm}.$

Hence, AD = $6\sqrt{3}$ cm.