A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(3pi)/8 #, and the triangle's area is #6 #. What is the area of the triangle's incircle?

1 Answer
Jun 28, 2017

The area of the incircle is #=1.89u^2#

Explanation:

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The area of the triangle is #A=6#

The angle #hatA=1/12pi#

The angle #hatB=3/8pi#

The angle #hatC=pi-(2/24pi+9/24pi)=13/24pi#

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(12/(sin(pi/12)*sin(3/8pi)*sin(13/24pi)))#

#=7.11#

Therefore,

#a=7.11sin(1/12pi)=1.84#

#b=7.11sin(3/8pi)=6.57#

#c=7.11sin(13/24pi)=7.05#

The radius of the incircle is #=r#

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

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

#=12/(15.5)=0.78#

The area of the incircle is

#area=pi*r^2=pi*0.78^2=1.89u^2#