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#