A triangle has vertices A, B, and C. Vertex A has an angle of $pi/8$ , vertex B has an angle of $(pi)/12$ , and the triangle's area is $1$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-02-01T14:53:26

Area of triangle’s Incircle $A_i = color(green)0.2424$

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Three angles are $A = pi / 8, B = pi / 12, C = pi - pi/8 - pi/12 = (19pi)/24$

Given Area of triangle A_t = 1#

$bc = A_t / ((1/2) sin A) = (2 * 1) / sin (pi / 8) = 5.2263$

$ca = A_t / ((1/2) sin B) = 2 / sin (pi / 12) = 7.7274$

$ab = A_t / ((1/2) sin C) = 2 / sin ((19pi)/24) = 3.2854$

$sqrt(ab * bc * ca)= (abc) = sqrt(5.2263 * 7.7274 * 3.2854) = 11.5188$

$a = (abc) / (bc) = 11.5188 / 5.2263 = 2.204$

$b = (abc) / (ca) = 11.5188 / 7.7274 = 1.4906$

$c = (abc) / (AB) = 11.5188 / 3.2854 = 3.5061$

Semi perimeter $s = (a + b + c) / 2 = (2.204 + 1.4906 + 3.5061) / 2 ~~ 3.6$

Inradius $I_r = A_t / (s/2) = 1 / 3.6$

Area of Incircle $A_i = pi (I_r)^2 = pi * (1/3.6)^2 = color(green)(0.2424)$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 pi/8 , vertex B has an angle of (pi)/12 (pi)/12 , and the triangle's area is 1 1 . What is the area of the triangle's incircle?