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?

Answer 1

`"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`