Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/8`, vertex B has an angle of `( pi)/4`, and the triangle's area is `18`. What is the area of the triangle's incircle?

Answer 1

`~~6.97`

Explanation:

The third angle of the triangle is `/_C = pi -pi/4-pi/8={5 pi}/8`

Sine law gives us

`a/{sin (pi/8)} = b/{sin(pi/4)}=c/{sin({5 pi}/8)} = k ( "say")`

The area is given by

`A = 1/2 b c sin/_A = 1/2 k^2 sin/_A sin/_B sin/_C = 1/2 k^2 sin(pi/8) sin(pi/4)sin({5pi}/8) = 1/2 k^2 sin(pi/8) sin(pi/4) cos(pi/4) = 1/4 k^2 sin^2(pi/4) = 1/8 k^2`

and thus `k^2 = 8 xx 18 = 144` or `k=12`.

Now, the radius of the incircle can be easily found from

`A = 1/2 (a+b+c)r`

which one can derive simply by joining each of the vertices with the incenter. So, we have

`1/2 k(sin(pi/8) +sin(pi/4)+sin({5pi}/8))r = 18`

and so

`r = 6/(sin(pi/8) +sin(pi/4)+sin({5pi}/8)) ~~ 1.49`

Thus the area of the incircle is

`pi r^2 ~~ 6.97`