Question

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?

Answer 1

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

Explanation:

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`