If CD = 10 m, the length of AB is :

1. $\dfrac{40}{\sqrt{3}}$ m

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

3. 100 m

4. 30 m

We know that,

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

From figure,

In △ BCD,

⇒ tan 30° = $\dfrac{BD}{CD}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{BD}{10}$

⇒ BD = $\dfrac{10}{\sqrt{3}}$ m.

In △ ACD,

⇒ tan 60° = $\dfrac{AD}{CD}$

⇒ $\sqrt{3} = \dfrac{AD}{10}$

⇒ AD = $10\sqrt{3}$ cm.

From figure,

AB = AD + BD = $10\sqrt{3} + \dfrac{10}{\sqrt{3}}$

$= \dfrac{10\sqrt{3} \times \sqrt{3} + 10}{\sqrt{3}} = \dfrac{30 + 10}{\sqrt{3}} = \dfrac{40}{\sqrt{3}}$ m.

Hence, Option 1 is the correct option.