Question

AB is 54.64
If true then enter $1$ and if false then enter $0$

Answer 1

In $△AFC$,
$\tan 45=\frac{FC}{AC}$
$1=\frac{20}{AC}$
$AC=20$
In $△EDB$,
$\tan 60=\frac{ED}{DB}$
$DB=\frac{30}{\sqrt{3}}$
$DB=10\sqrt{3}$
Draw a perpendicular from F on ED to meet at M,
$EM=ED-FC$
$EM=30-20=10$
Now In $△FEM$,
$\tan 60=\frac{FM}{EM}$
$FM=10\sqrt{3}$
Thus, $AB=AC+DB+FM$
$AB=20+10\sqrt{3}+10\sqrt{3}$
$AB=54.64$