A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/3 #, and the triangle's area is #18 #. What is the area of the triangle's incircle?

Given that in #\Delta ABC#, #A=\pi/8#, #B={\pi}/3#

#C=\pi-A-B#

#=\pi-\pi/8-{\pi}/3#

#={13\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/8)}=\frac{b}{\sin ({\pi}/3)}=\frac{c}{\sin ({13\pi}/24)}=k\ \text{let}#

#a=k\sin(\pi/8)=0.383k#

# b=k\sin({\pi}/3)=0.866k#

#c=k\sin({13\pi}/24)=0.991k#

#s=\frac{a+b+c}{2}#

#=\frac{0.383k+0.866k+0.991k}{2}=1.12k#

Area of #\Delta ABC# from Hero's formula

#\Delta=\sqrt{s(s-a)(s-b)(s-c)}#

#18=\sqrt{1.12k(1.12k-0.383k)(1.12k-0.866k)(1.12k-0.991k)}#

#18=0.1644k^2#

#k^2=109.4506#

Now, the in-radius (#r#) of #\Delta ABC#

#r=\frac{\Delta}{s}#

#r=\frac{18}{1.12k}#

Hence, the area of inscribed circle of #\Delta ABC#

#=\pi r^2#

#=\pi (18/{1.12k})^2#

#=\frac{324\pi}{1.2544k^2}#

#=\frac{811.4445}{109.4506}\quad (\because k^2=109.4506)#

#=7.414\ \text{unit}^2#

#hat A = pi/8, hat B = pi/3, hat C = (13pi)/24, A_t = 18#

#A_t = 1/2 ab sin C = 1/2 bc sin A = 1/2 ca sin B#

#ab = (2 A_t) / sin C = 36 / sin ((13pi)/24) ~~ 36.3106#

#bc = (2 A_t) / sin A = 36 / sin ((pi)/8) ~~ 94.0725#

#ca = (2 A_t) / sin B = 36 / sin ((pi)/3) ~~ 41.5692#

#a = sqrt (ab * bc * ca) / (bc) = sqrt ((ab * ca) / (bc))#

#a = sqrt((36.3106*41.5692)/94.0725) ~~ 4#

Similarly, #b = sqrt((36.3106 * 94.0725)/41.5692) ~~ 9.0649#

#c = sqrt((94.0725*41.5692)/36.3106) ~~ 10.3777#

Semi perimeter #s = (a + b + c) / 2 = 23.4426/2 = 11.7213#

Incircle radius #r = A_t / s = 18 / 11.7213 ~~ 1.5357#

Area of Incircle #A_i = pi r^2 = pi * (1.5357)^2 ~~ 7.4087#