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 #4 #. What is the area of the triangle's incircle?

Question
Answer 1

The area of the triangle's incircle is #1.07# sq.unit.

#/_A = pi/6= 180/6=30^0 , /_B = pi/12=180/12= 15^0#

# /_C= 180-(30+15)=135^0 ;A_t=4# We know Area ,

# A_t= 1/2*b*c*sin A or b*c=(2*4)/sin 30 = 16 #, similarly ,

#a*c=(2*4)/sin 15 ~~ 30.91 #, and #a*b=(2*4)/sin 135~~ 11.31 #

#(a*b)*(b*c)*(c.a)=(a b c)^2= (16*30.91*11.31) # or

#a b c=sqrt(5595.31) = 74.80 #

#a= (a b c)/(b c)=74.80/16~~4.68#

#b= (a b c)/(ac)=74.80/30.91~~2.42#

#c= (a b c)/(ab)=74.80/11.31~~6.61#

Semi perimeter : #S/2=(4.68+2.42+6.61)/2~~6.855#

Incircle radius is #r_i= A_t/(S/2) = 4/6.855~~0.58#

Incircle Area = #A_i= pi* r_i^2= pi*0.58^2 ~~1.07# sq.unit

The area of the triangle's incircle is #1.07# sq.unit [Ans]

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