Find AB and BC, if :

Let BC be x cm

BD = BC + CD = x + 20 cm

In Δ ABD,

$\text{tan 30°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{AB}{BD}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{AB}{x + 20}\\[1em] ⇒ x + 20 = AB\sqrt3 …………..(1)$

In Δ ABC,

$\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{AB}{BC}\\[1em] ⇒ \sqrt3 = \dfrac{AB}{x}\\[1em] ⇒ x = \dfrac{AB}{\sqrt3} …………..(2)$

From equation (1),

$\dfrac{AB}{\sqrt3} + 20 = AB\sqrt3\\[1em] ⇒ AB + 20\sqrt3 = 3AB\\[1em] ⇒ 34.64 = 3AB - AB\\[1em] ⇒ AB = \dfrac{34.64}{2}\\[1em] ⇒ AB = 17.32$

From equation (2),

x = $\dfrac{AB}{\sqrt3}$

x = $\dfrac{17.32}{\sqrt3}$

x = 10 cm

Hence, AB = 17.32 cm and BC = 10 cm.