Find AD, if :

Let AC = x. Therefore, BC = x.

In Δ ADC,

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

And,

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

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

In Δ ABD, according to Pythagoras theorem,

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

$⇒ 100^2 = \Big(\dfrac{3x}{2}\Big)^2 + \Big(\dfrac{\sqrt3x}{2}\Big)^2 \\[1em] ⇒ 10000 = \dfrac{9x^2}{4} + \dfrac{3x^2}{4} \\[1em] ⇒ 10000 = \dfrac{9x^2 + 3x^2}{4}\\[1em] ⇒ 10000 = \dfrac{12x^2}{4}\\[1em] ⇒ 10000 = 3x^2\\[1em] ⇒ x^2 = \dfrac{10000}{3}\\[1em] ⇒ x = \sqrt\dfrac{10000}{3}\\[1em] ⇒ x = \dfrac{100}{\sqrt3}$

Substituting the value of x in AD,

$AD = \dfrac{x\sqrt3}{2}\\[1em] ⇒ AD = \dfrac{\dfrac{100}{\sqrt3} \times \sqrt3}{2}\\[1em] ⇒ AD = \dfrac{\dfrac{100}{\cancel {\sqrt3}} \times \cancel {\sqrt3}}{2}\\[1em] ⇒ AD = 50 m$

Hence, AD = 50 m.