A triangle has vertices A, B, and C. Vertex A has an angle of $pi/12$ , vertex B has an angle of $pi/6$ , and the triangle's area is $9$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-01-23T21:54:39

$Area ~~ 2.40$

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Given $angleA = pi/12, angle B = pi/6$ , and the area, $Delta = 9$

Compute $angle C$ :

$angle C = pi - angleA - angleB$

$angle C = pi - pi/12 - pi/6$

$angle C = (3pi)/4$

From the reference, the square of radius of the incircle is:

$r^2 = Delta/(cot(A/2) + cot(B/2) + cot(C/2))$

The area of the incircle is:

$Area = (piDelta)/(cot(A/2) + cot(B/2) + cot(C/2))$

Substitute the values for, $A, B, C, and Delta$

$Area = (9pi)/(cot(pi/24) + cot(pi/12) + cot((3pi)/8))$

$Area ~~ 2.40$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 pi/12 , vertex B has an angle of pi/6 pi/6 , and the triangle's area is 9 9 . What is the area of the triangle's incircle?