A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/8 #, and the triangle's area is #4 #. What is the area of the triangle's incircle?

The distances from the incenter to each side are equal to the inscribed circle's radius. The area of the triangle is equal to 12×r×(the triangle's perimeter), 1 2 × r × ( the triangle's perimeter ) , where r is the inscribed circle's radius.

First to find the perimeter of the triangle.

#hat A = pi/12, hat B = pi/8, hat C = (19pi)/24, A_t = 4#

Area of triangle #A_t = (1/2)ab sin C = (1/2)bc sin A = (1/2)ca sin B#

#ab = (4 * 2) / sin ((19)/24) = 13.14#

Similarly, #bc = (4 * 2) / sin (pi/12) = 30.91#

#ca = (4 * 2) / sin (pi/8) = 20.91#

#ab * bc * ca = (abc)^2 = 13.14 * 30.91 * 20.91#

#abc = sqrt(13.14 * 30.91 * 20.91)#

#a = (abc) / (bc) = sqrt(13.14 * 30.91 * 20.91) / 30.91 = 2.98#

Likewise, #b = (abc) / (ca) = sqrt(13.14 * 30.91 * 20.91) / 20.91 = 4.41#

#c = (abc) / (ab) = sqrt(13.14 * 30.91 * 20.91) / 13.14 = 7.01#

#"Perimeter of the triangle " p = a + b + c = 2.98 + 4.41 + 7.01 = 14.4#

#"Radius of incircle " r = A_t / (12 * p) = 4 / (12 * 14.4) = 0.023#

#color(purple)("Area of incircle " A_r = pi r^2 = pi * 0.023^2 = 0.00166#