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?

Question

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?

Mason Adams · Accepted Answer

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

Jayden Jackson · Answer

undefined The area of the triangle's incircle is $ \frac{15\sqrt{3}}{6} $, which simplifies to $ \frac{5\sqrt{3}}{2} $.