A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #(5pi)/12 #, and the triangle's area is #28 #. What is the area of the triangle's incircle?

1 Answer
Narad T.
Jul 1, 2017

The area of the incircle is #=11.78u^2#

Explanation:

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The area of the triangle is #A=28#

The angle #hatA=1/8pi#

The angle #hatB=5/12pi#

The angle #hatC=pi-(1/8pi+5/12pi)=11/24pi#

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(56/(sin(1/8pi)*sin(5/12pi)*sin(11/24pi)))#

#=12.36#

Therefore,

#a=12.36sin(1/8pi)=4.73#

#b=12.36sin(5/12pi)=11.94#

#c=12.36sin(11/24pi)=12.24#

The radius of the incircle is #=r#

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

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

#=56/(28.91)=1.94#

The area of the incircle is

#area=pi*r^2=pi*1.94^2=11.78u^2#

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