Find AB and BC, if :

Let BC be x cm

BD = BC + CD = x + 20 cm

In Δ ABD,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{AB}{BD}\\[1em] ⇒ 1 = \dfrac{AB}{x + 20}\\[1em] ⇒ x + 20 = AB …………..(2)$

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),

$\dfrac{AB}{\sqrt3} + 20 = AB\\[1em] ⇒ AB + 20\sqrt3 = \sqrt3AB\\[1em] ⇒ 34.64 = \sqrt3AB - AB\\[1em] ⇒ 34.64 = AB(1.73 - 1)\\[1em] ⇒ 34.64 = AB \times (0.73)\\[1em] ⇒ AB = \dfrac{34.64}{0.73}\\[1em] ⇒ AB = 47.32$

From equation (2),

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

x = $\dfrac{47.32}{\sqrt3}$

x = 27.32 cm

Hence, AB = 47.32 cm and BC = 27.32 cm.