A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{3}$ $pi/3$ , vertex B has an angle of $\frac{5 π}{12}$ $(5 pi)/12$ , and the triangle's area is $27$ $27$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-07-29T10:56:38

$Area of inscribed circle = A_{i} = π r^{2} \approx 16.0643$ $color(indigo)("Area of inscribed circle " = A_i = pi r^2 ~~ 16.0643$

$"Area of "Delta = A_t = (1/2) a b sin C = (1/2) b c sin A = (1/2) c a sin B$

$"Given " hat A = pi/3, hat B = (5pi)/12, hat C = pi/4$ , A_t = 27#

![http://mathibayon.blogspot.com/2015/01/ derivation-of-formula-for-radius-of-incircle.html

$a b = (2 A_t) / sin C = 54 / sin (pi/4) ~~ 76.3675$

$b c = (2 A_t) / sin A = 54 / sin (pi/3) = 62.3538$

$c a = (2 A_t) / sin B = 54 / sin ((5pi)/12) = 55.9049$

$a = (a b c) / (b c) = sqrt(76.3675 * 62.3538 * 55.9049) / 62.3538 = 7.8948$

$b = (a b c) / (c a) = sqrt(76.3675 * 62.3538 * 55.9049) / 55.9049 = 9.2291$

$c = (a b c) / (a b) = sqrt(76.3675 * 62.3538 * 55.9049) / 76.3675= 6.7562$

$"Semi-perimeter " = s = (a + b + c) / 2 = 23.8801 / 2 = 11.9401$

$"Radius of inscribed circle " = r = A_t / s = 27 / 11.9401$

$color(indigo)("Area of inscribed circle " = A_i = pi r^2 = pi * (27/ 11.9401)^2 ~~ 16.0643$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/3 π3pi/3 , vertex B has an angle of (5 pi)/12 5π12(5 pi)/12 , and the triangle's area is 27 2727 . What is the area of the triangle's incircle?