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#