Find AB.

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Solution

Drop a perpendicular $F X$ from $F$ to $E D$ .
$E X = E D - F C = 30 - 20 = 10$
In $△ F X E$ ,
$F X = (E X) tan 60^{\circ} F X = 10 \sqrt{3} F X = C D = 10 \sqrt{3}$
In $△ A C F$ ,
$A C = 20 cot 45^{\circ} A C = 20 (1) = 20$
In $△ B D E$ ,
$B D = (D E) cot 60^{\circ} = \frac{30}{\sqrt{3}} = 10 \sqrt{3}$
Now, $A B = A C + C D + D B = 20 + 10 \sqrt{3} + 10 \sqrt{3} = 20 + 20 \sqrt{3}$
$A B = 20 (1 + \sqrt{3})$

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