Question

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?

Answer 1

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

Explanation:

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`