A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(5pi)/8 #, and the triangle's area is #19 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-26T17:47:43

#5.7975\ \text{unit}^2#

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#