A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/6 #, and the triangle's area is #4 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-07-11T11:50:01

#1.073\ \text{unit}^2#

Given that in #\Delta ABC#, #A=\pi/12#, #B=\pi/6#

#C=\pi-A-B#

#=\pi-\pi/12-\pi/6#

#={3\pi}/4#

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 (\pi/6)}=\frac{c}{\sin ({3\pi}/4)}=k\ \text{let}#

#a=k\sin(\pi/12)=0.259k#

# b=k\sin(\pi/6)=0.5k#

#c=k\sin({3\pi}/4)=0.707k#

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

#=\frac{0.259k+0.5k+0.707k}{2}=0.733k#

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

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

#4=\sqrt{0.733k(0.733k-0.259k)(0.733k-0.5k)(0.733k-0.707k)}#

#4=0.0459k^2#

#k^2=87.187#

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

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

#r=\frac{4}{0.733k}#

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

#=\pi r^2#

#=\pi (4/{0.733k})^2#

#=\frac{16\pi}{0.537289k^2}#

#=\frac{93.5539}{87.187}\quad (\because k^2=87.187)#

#=1.073\ \text{unit}^2#