Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `(2pi)/3`, and the triangle's area is `8`. What is the area of the triangle's incircle?

Answer 1

Incircle area `=2.38(2dp)`sq.unit

Explanation:

Angles of the vertices are `/_A=pi/12=180/12=15^0; /_B=2*pi/3=120^0 ; /_C=180-(120=15)=45^0`
Sides `(A*C*sin120)/2=8 or A*C=16/sin120=18.48;`
`(B*C*sin15)/2=8 or B*C=16/sin15=61.82;`
`(A*B*sin45)/2=8 or A*B=16/sin45=22.63;`
`A*B*C=sqrt(A*C*B*C*A*B)=sqrt(18.48*61.82*22.63)=160.79`
Side `A=160.79/61.82=2.60; B=160.79/18.48=8.7;C=160.79/22.63=7.11`
Semi perimeter;`s=(A+B+C)/2=(2.6+8.7+7.1)/2=9.2`
Incircle radius; i.e (area of triangle/semi perimeter) `r=8.0/9.2=0.87`
Incircle area `A_i=pi*0.87^2=2.38(2dp)`sq.unit[Ans]