In the given triangle, the length of AB is :

1. 160 cm

2. 120 cm

3. 60 cm

4. 40 cm

Let AC = x. Therefore, DC = x.

In Δ ABC,

$\text{cos 60°} = \dfrac{Base}{Hypotenuse}\\[1em] ⇒ \dfrac{1}{2} = \dfrac{BC}{x}\\[1em] ⇒ BC = \dfrac{x}{2}$

$\text{sin 60°} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \dfrac{\sqrt3}{2} = \dfrac{AB}{x}\\[1em] ⇒ AB = \dfrac{x\sqrt3}{2}$

$BD = BC + CD\\[1em] ⇒ BD = \dfrac{x}{2} + x\\[1em] ⇒ BD = \dfrac{x + 2x}{2}\\[1em] ⇒ BD = \dfrac{3x}{2}$

In Δ ABD, according to Pythagoras theorem,

⇒ AD² = BD² + AB² (∵ AD is hypotenuse)

$⇒ 80^2 = \Big(\dfrac{3x}{2}\Big)^2 + \Big(\dfrac{\sqrt3x}{2}\Big)^2 \\[1em] ⇒ 6400 = \dfrac{9x^2}{4} + \dfrac{3x^2}{4} \\[1em] ⇒ 6400 = \dfrac{9x^2 + 3x^2}{4}\\[1em] ⇒ 6400 = \dfrac{12x^2}{4}\\[1em] ⇒ 6400 = 3x^2\\[1em] ⇒ x^2 = \dfrac{6400}{3}\\[1em] ⇒ x = \sqrt\dfrac{6400}{3}\\[1em] ⇒ x = \dfrac{80}{\sqrt3}$

Substituting the value of x in AB,

$AB = \dfrac{x\sqrt3}{2}\\[1em] ⇒ AB = \dfrac{\dfrac{80}{\sqrt3} \times \sqrt3}{2}\\[1em] ⇒ AB = \dfrac{\dfrac{80}{\cancel {\sqrt3}} \times \cancel {\sqrt3}}{2}\\[1em] ⇒ AB = 40 m$

Hence, option 4 is the correct option.