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 $16$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-01-09T10:56:33

Area of I’m circle $color(green)(A_i = 9.19) sq. Units$

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Given $A = pi /3, B = (5pi) / 12, C = pi - A - B = pi / 4$

Area of triangle $A_t = 16$

$a*b = A_t / ((1/2) sin C) = 16 / (0.5 sin (pi/3)) ~~ 36.95$

Similarly,
$b*c = 16 / (0.5 sin A) = 16 / (0.5 * sin ((5pi)/12)) ~~ 33.13$

$c*a = 16 / (0.5 sin B) = 16 / (0.5 * sin (pi/4)) ~~ 45.25$

$(ab * bc * ca) = (abc)^2 = 36.95 * 33.13 * 45.25 = 55392.95$

$abc = sqrt(55392.95) ~~ 235 36$

$a = (abc) / (bc) = 235.36 / 33.13 ~~ 7.1$

$b = (abc) / (ca) = 235.36 / 45.25 ~~ 5.2$

$c = (abc) / (ab) = 235.36 / 36.95 ~~ 6.37$

Semi perimeter $S/ 2 = (a + b + c) / 2 = (7.1 + 5.2 + 6.37) / 2 = 9.34$

In radius $r = A_t / (S/2) = 16 / 9.34 = 1.71$

Area of in circle $A_i = pi r^2 = pi * (1.71)^2 = color(green)(9.19)$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/3 pi/3 , vertex B has an angle of (5 pi)/12 (5 pi)/12 , and the triangle's area is 16 16 . What is the area of the triangle's incircle?