Length of AB is equal to:

1. $20\sqrt{3}$ cm

2. $\dfrac{20}{\sqrt{3}}$ cm

3. 20 cm

4. none of these

From figure,

DC = CA = x (let)

By formula,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

In triangle ABD,

$\Rightarrow \text{tan D} = \dfrac{AB}{DB} \\[1em] \Rightarrow \text{tan 30°} = \dfrac{AB}{DC + CB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{x + 20} \\[1em] \Rightarrow AB = \dfrac{x + 20}{\sqrt{3}}.$

Since ΔABC is a right angled triangle, using pythagoras theorem,

⇒ Hypotenuse² = Base² + Height²

⇒ AC² = CB² + AB²

$\Rightarrow x^2 = 20^2 + \Big(\dfrac{x + 20}{\sqrt{3}}\Big)^2\\[1em] \Rightarrow x^2 = 400 + \Big(\dfrac{x^2 + 400 + 40x}{3}\Big)\\[1em] \Rightarrow x^2 = \Big(\dfrac{1200 + x^2 + 400 + 40x}{3}\Big)\\[1em] \Rightarrow 3x^2 = 1200 + x^2 + 400 + 40x\\[1em] \Rightarrow 3x^2 = 1600 + x^2 + 40x\\[1em] \Rightarrow 3x^2 - 1600 - x^2 - 40x = 0\\[1em] \Rightarrow 2x^2 - 40x - 1600 = 0\\[1em] \Rightarrow 2(x^2 - 20x - 800) = 0 \\[1em] \Rightarrow x^2 - 20x - 800 = 0\\[1em] \Rightarrow x^2 - 40x + 20x - 800 = 0\\[1em] \Rightarrow x(x - 40) + 20(x - 40) = 0\\[1em] \Rightarrow (x - 40)(x + 20) = 0\\[1em] \Rightarrow (x - 40) = 0 \text{ or } (x + 20) = 0\\[1em] \Rightarrow x = 40 \text{ or } x = -20.$

As length of side of triangles cannot be negative. So, x = 40 cm.

DC = AC = x = 40 cm.

$AB = \dfrac{x + 20}{\sqrt{3}} = \dfrac{40 + 20}{\sqrt{3}} = \dfrac{60}{\sqrt{3}} = 20\sqrt{3}$ cm.

Hence, option 1 is the correct option.