Given that in \Delta ABC, A=\pi/12, B={5\pi}/8
C=\pi-A-B
=\pi-\pi/12-{5\pi}/8
={7\pi}/24
from sine in \Delta ABC, we have
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
\frac{a}{\sin(\pi/12)}=\frac{b}{\sin ({5\pi}/8)}=\frac{c}{\sin ({7\pi}/24)}=k\ \text{let}
a=k\sin(\pi/12)=0.259k
b=k\sin({5\pi}/8)=0.924k
c=k\sin({\pi}/24)=0.793k
s=\frac{a+b+c}{2}
=\frac{0.259k+0.924k+0.793k}{2}=0.988k
Area of \Delta ABC from Hero's formula
\Delta=\sqrt{s(s-a)(s-b)(s-c)}
19=\sqrt{0.988k(0.988k-0.259k)(0.988k-0.924k)(0.988k-0.793k)}
19=0.09481k^2
k^2=200.4029
Now, the in-radius (r) of \Delta ABC
r=\frac{\Delta}{s}
r=\frac{19}{0.988k}
Hence, the area of inscribed circle of \Delta ABC
=\pi r^2
=\pi (19/{0.988k})^2
=\frac{361\pi}{0.976144k^2}
=\frac{1161.8316}{200.4029}\quad (\because k^2=200.4029)
=5.7975\ \text{unit}^2
