Find AB.

Draw a line FP parallel to CD such that FP = CD and FC = PD.

In Δ ACF,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{20}{AC}\\[1em] ⇒ AC = 20$

In Δ DEB,

$\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{30}{BD}\\[1em] ⇒ BD = \dfrac{30}{\sqrt3}\\[1em] ⇒ BD = \dfrac{30 \times \sqrt3}{\sqrt3 \times \sqrt3}\\[1em] ⇒ BD = \dfrac{30\sqrt3}{3}\\[1em] ⇒ BD = 10\sqrt3 = 17.32$

As, FC = 20 and ED = 30

EP = ED - PD = ED - FC = 30 - 20 = 10 cm

In Δ EFP,

$\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{FP}{10}\\[1em] ⇒ FP = 10 \sqrt3 = 17.32$

Thus AB = AC + CD + BD

= 20 + 17.32 + 17.32

= 54.64

Hence, AB = 54.64.