Question

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?

Answer 1

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

Explanation:

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`