A triangle has vertices A, B, and C. Vertex A has an angle of #pi/6 #, vertex B has an angle of #(pi)/12 #, and the triangle's area is #15 #. What is the area of the triangle's incircle?

1 Answer
Feb 18, 2018

#"Area of the triangle's incircle "=87.934#

Explanation:

#"Given:"#

#"Angle"theta_A=pi/6#
#"Angle"theta_B=pi/12#
#"Area of triangle"=15#

#"To find:"#

#"Area of the triangle's incircle"#

#theta_A+theta_B+theta_C=pi#

#theta_A+theta_B=pi/6+pi/12#

#pi/6+pi/12=pi/4#

#theta_A+theta_B=pi/4#

#pi/4+theta_C=pi#

#theta_C=pi-pi/4#

#pi-pi/4=(3pi)/4#

#theta_C=(3pi)/4#

#"Let " r " be the radius of the triangle's incircle"#

#"Area of the triangle's incircle"=pir^2#

#"Vertex Angle at A is "theta_A=pi/6#

#"Semi-Vertex angle at A is "=pi/12#

#"tangent length at vertex A is "l_A=rtan(pi/12)#

#"Vertex Angle at B is "theta_B=pi/12#

#"Semi-Vertex angle at B is "=pi/24#

#"tangent length at vertex B is "l_B=rtan(pi/24)#

#"Vertex Angle at C is "theta_C=(3pi)/4#

#"Semi-Vertex angle at C is "=(3pi)/8#

#"tangent length at vertex C is "l_C=rtan((3pi)/8)#

#"Length of side AB is "=l_A+l_B=rtan(pi/12)+rtan(pi/24)#

#"Length of side AB is "=r(tan(pi/12)+tan(pi/24))#

#"Length of side AC is "=l_A+l_C=rtan(pi/12)+rtan((3pi)/8)#

#"Length of side AC is "=r(tan(pi/12)+tan((3pi)/8))#

#"Vertex Angle at A is "A=pi/6#

#"Area of triangle is "A=15#

#"Area of triangle "=1/2xxABxxACxxsintheta_A#

#"Substituting"#

#15=1/2xx(r(tan(pi/12)+tan(pi/24)))xx(r(tan(pi/12)+tan((3pi)/8)))#

#15=r^2/2xx(tan(pi/12)+tan(pi/24))(tan(pi/12)+tan((3pi)/8))#

#r^2/2xx(tan(pi/12)+tan(pi/24))(tan(pi/12)+tan((3pi)/8))=15#

#r^2/2=15/((tan(pi/12)+tan(pi/24))(tan(pi/12)+tan((3pi)/8)))#

#"Area of the triangle's incircle"=pir^2#

#r^2=30/((tan(pi/12)+tan(pi/24))(tan(pi/12)+tan((3pi)/8)))#

#"Area of the triangle's incircle"=pixx30/((tan(pi/12)+tan(pi/24))(tan(pi/12)+tan((3pi)/8)))#

#"Area of the triangle's incircle"=(30pi)/((tan(pi/12)+tan(pi/24))(tan(pi/12)+tan((3pi)/8)))#

#"Area of the triangle's incircle "=87.934#