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 `8`. What is the area of the triangle's incircle?

Answer 1

`A_(o.) = pi*r_Delta^2 =[8/(8+4/sqrt(2-sqrt(3)))]^2*pi`

`A_(o.) ~~ [8/(8+7.73)]^2pi~~.259pi`

Explanation:

Firs by AA the triangle is isosceles triangle with sides `a=b` see geometric figure. Thus using the sine law and the cross product we can write:
1) `a/sinalpha = b/sinbeta = c/singamma`
2) `A_Delta =1/2 |a|*|c|*sin(alpha)= 1/2(ac)sin(alpha)`
3) `a=b " and " alpha = beta = pi/12`

From 1) we have `a = sinalpha/singamma c`
Substitute into 2) `2A_Delta=sin^2alpha/sinbeta c^2`
Now letting `A_Delta=8; sinalpha=sin(pi/12); sinbeta= sin(5/6pi)`
`16=sin^2(pi/12)/sin(5/6pi) c^2`
`c=sqrt(16*sin(5/6pi)/sin^2(pi/12))=(4)/sin(pi/12)sqrt(sin(5/6pi))`
`c=(4*2)/sqrt((2-sqrt(3)))color(red)(sqrt(sin(5/6pi))`
`color(red)(sqrt(sin(5/6pi)) = color(blue)(cos(pi/2-5/6pi)=cos(pi/3)=1/2)`
`c=(4*2)/sqrt((2-sqrt(3)))color(blue)(1/2)=(4)/sqrt(2-sqrt(3) )`
`a= sin(pi/12)/cos(pi/3)*(4)/sqrt(2-sqrt(3) ) = 4 =b`
Now the task is to find the radius of the Incenter, Inradius `r_Delta`
`r_Delta = A_Delta/P_Delta` Where
`A_Delta` = Area and `P_Delta` = Perimeter
`r_Delta = 8/(8+(4)/sqrt(2-sqrt(3))`
An the `A_(o.) = pi*r_Delta^2 =[8/(8+4/sqrt(2-sqrt(3)))]^2pi`
`A_(o.) ~~ [8/(8+7.73)]^2pi~~.259pi`