Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `pi/8`, and the triangle's area is `9`. What is the area of the triangle's incircle?

Answer 1

`color(indigo)("Area of incircle " = pi r^2 = pi * (0.8318)^2 = 2.174`

Explanation:

`hat A = pi/12, hat B pi/8, hat C = (19pi) / 24), A_t = 9`

`A_t = (1/2) ab sin C = (1/2(bc sin A = (1/2) ac sin B`

`ab =( 2A_t) /sin C = (2 * 9) / sin ((19pi)/24) = 29.57`

`bc = (2 * 9) / sin (pi/12) = 69.55`

`ca = (2 * 9) / sin (pi/8) = 47 04`

`a = sqrt(abc)^2 / (bc) =sqrt (29.57 * 69.55 * 47.04) / 69.55 = 4.48`

`b = sqrt(abc)^2 / (ac) =sqrt (29.57 * 69.55 * 47.04) / 47.04 = 6.62`

`c = sqrt(abc)^2 / (ab) =sqrt (29.57 * 69.55 * 47.04) / 29.57 = 10.54`

`"Semiperimeter " = s = (a = b + c) / 2 = (4.48 + 6.62 + 10.54) / 2 = 10.82`

`color(chocolate)("Incircle radius " = r = A_t / s = 9 / 10.82 = 0.8318`

`color(indigo)("Area of incircle " = pi r^2 = pi * (0.8318)^2 = 2.174`