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 $12$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-06-20T06:05:54

$color(indigo)("Area of incircle " = A_i = pi * (12/15.5)^2 = 1.88 " sq units"$

![ jwilson.coe.uga.edu )

$hat A = pi/6, hat B = pi/12, hat C = (3pi) / 4, A_t = 12$

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

$ab = (2 A_t) / sin C = 24 / sin ((3pi)/4) = 33.94$

$bc = 24 / sin A = 24/ sin (pi/6) = 48$

$ca = 24 / sin C = 24 / sin(pi/12) = 92.73$

$a = (abc) / (bc) = sqrt(33.94 * 48 * 92.73) / 48 = 8.1$

$b = (abc) / (ac) = sqrt(33.94 * 48 * 92.73) / 92.73 = 4.19$

$c = (abc) / (ab) = sqrt(33.94 * 48 * 92.73) / 33.94 = 11.45$

$"Semiperimeter " = s = (a + b + c) / 2 = (8.1 + 4.19 + 11.45) / 2 = 15.5$

$"Incircle radius " = r = A_t / s = 12 / 15.5$

$color(indigo)("Area of incircle " = A_i = pi * (12/15.5)^2 = 1.88 " sq units"$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/6 pi/6 , vertex B has an angle of (pi)/12 (pi)/12 , and the triangle's area is 12 12 . What is the area of the triangle's incircle?