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?

1 Answer
Binayaka C.
Aug 13, 2016

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]

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