A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(5pi)/8 #, and the triangle's area is #13 #. What is the area of the triangle's incircle?

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Answer 1 · 2018-06-21T04:36:54

#color(purple)("Area of incircle " = A_i = pi * (13/23.12)^2 = 0.9933 " sq units"#

#hat A = pi/12, hat B = (5pi)/8, hat C = (7pi) / 24, A_t = 13#

#A_t = (1/2) a b sin C = (1/2) bc sin A = (1/2) ca sin B#

#ab = (2 A_t) / sin C = 26 / sin ((7pi)/24) = 32.77#

#bc = A_t / sin A = 26/ sin (pi/12) = 100.46#

#ca = A_t / sin C = 26 / sin(5pi)/8) = 28.14#

#a = (abc) / (bc) = sqrt(32.77 * 100.46 * 28.1) / 100.46 = 3.03#

#b = (abc) / (ac) = sqrt(32.77 * 100.46 * 28.1) / 28.14 =10.81#

#c = (abc) / (ab) = sqrt(32.77 * 100.46 * 28.1) / 32.77 = 9.28#

#"Semiperimeter " = s = (a + b + c) / 2 = (3.03 + 10.81 + 9.28) / 2 = 11.56#

#"Incircle radius " = r = A_t / s = 13 / 23.12#

#color(purple)("Area of incircle " = A_i = pi * (13/23.12)^2 = 0.9933 " sq units"#