A triangle has vertices A, B, and C. Vertex A has an angle of #pi/3 #, vertex B has an angle of #(5 pi)/12 #, and the triangle's area is #4 #. What is the area of the triangle's incircle?

firstly, you have to look for the area of triangle..

#"Area of a triangle" = l^2#

Where, #l = 4#

#"Area of a triangle" = 4^2#

#"Area of a triangle" = 16#square units

Let;

#A = pi/3, B = (5pi)/12#

Since; #A + B + C = pi#

Therefore; #C = pi - A - B#

#C = pi - pi/3 - (5pi)/12#

#C = (12pi - 4pi - 5pi)/12#

#C = (3pi)/12#

#C = pi/4#

Finding their respective angles..

#a xx b = "Area"/(1/2 xx sin C) = 16/(0.5 xx sin (pi/3)) = 36.95#

similarly..

#b xx c = "Area"/(1/2 xx sin A) = 16/(0.5 xx sin ((5pi)/12)) = 33.13#

similarly..

#c xx a = "Area"/(1/2 xx sin B) = 16/(0.5 xx sin (pi/4)) = 45.25#

#(ab xx bc xx ca) = 36.95 xx 33.13 xx 45.25#

#(abc)^2 = 55,392.946#

squareroot both sides..

#sqrt((abc)^2) = sqrt(55,392.946)#

#abc = 235.357#

Now to get #a#

divide both sides by #bc# and its value..

#(abc)/(bc) = 235.357/33.13#

#a = 71#

smilarly..

#abc = 235.357#

divide both sides by #ca# and its value

#(abc)/(ca) = 235.357/45.25#

#b = 5.2#

smilarly..

#abc = 235.357#

divide both sides by #ab# and its value..

#(abc)/(ab) = 235.357/36.95#

#c = 6.37#

For half a perimeter; #(a + b + c)/2 = (7.1 + 5.2 + 6.37)/2 = 9.34#

#"radius" (r) = "Area"/"half of perimeter" = A/"semi perimeter" = 16/9.34 = 1.71#

#"Area of a circle" = pir^2#

#"Area of a circle" = 3.142 xx (1.71)^2#

#"Area of a circle" = 3.142 xx 2.9241#

#"Area of a circle" = 9.1875#

#"Area of a circle" = 9.19 "square units"#