A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/12 #, and the triangle's area is #7 #. What is the area of the triangle's incircle?

1 Answer
Narad T.
Jul 4, 2017

The area of the incircle is #=1.42u^2#

Explanation:

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The area of the triangle is #A=7#

The angle #hatA=1/12pi#

The angle #hatB=1/12pi#

The angle #hatC=pi-(1/12pi+1/12pi)=5/6pi#

The sine rule is

#a/sinA=b/sinB=c/sinC=k#

So,

#a=ksinA#

#b=ksinB#

#c=ksinC#

Let the height of the triangle be #=h# from the vertex #A# to the opposite side #BC#

The area of the triangle is

#A=1/2a*h#

But,

#h=csinB#

So,

#A=1/2ksinA*csinB=1/2ksinA*ksinC*sinB#

#A=1/2k^2*sinA*sinB*sinC#

#k^2=(2A)/(sinA*sinB*sinC)#

#k=sqrt((2A)/(sinA*sinB*sinC))#

#=sqrt(14/(sin(1/12pi)*sin(1/12pi)*sin(5/6pi)))#

#=20.44#

Therefore,

#a=20.44sin(1/12pi)=5.29#

#b=20.44sin(1/12pi)=5.29#

#c=20.44sin(11/24pi)=10.22#

The radius of the incircle is #=r#

#1/2*r*(a+b+c)=A#

#r=(2A)/(a+b+c)#

#=14/(20.8)=0.67#

The area of the incircle is

#area=pi*r^2=pi*0.67^2=1.42u^2#

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