Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `pi/4`, and the triangle's area is `3`. What is the area of the triangle's incircle?

Answer 1

`A_(@) = pir^2 = pi(0.52868)^2 = .2795pi`

Explanation:

Given a triangle with vertices A, B and C the area of triangle can be computed using the formula, `A_Delta= 1/2 bar(AC)*bar(AB) sinalpha` where `alpha` is the agnle between the two side `bar(AC)` and `bar(AB)`. We also should note the Area of the Triangle
can be computed using,
`A_Delta= 1/2 r (bar(AC)+bar(AB)+bar(BC))= 1/2rP; P= Perimeter " and " r` the radius of the inscribed circle.
We have 3 equations"
`A_Delta =3 = 1/2 bar(AC)*bar(AB) sin(pi/12)` ===> (1)
`A_Delta =3= 1/2 bar(AB)*bar(BC) sin(pi/4)` ====> (2)
`A_Delta =3= 1/2 bar(AC)*bar(BC) sin((8pi)/12)` ===> (3)

`bar(AC)/bar(BC) = sin(pi/4)/(sin(pi/12))` ====> (4)

`bar(AB)/bar(BC) = sin((8pi)/12)/(sin(pi/12))` ===> (5)

`bar(AB)/bar(AC) = sin((8pi)/12)/(sin(pi/4))` ===> (6)

Using (1) and (6) we write:

`3 = 1/2 bar(AC)*bar(AC) sin((8pi)/12)/(sin(pi/4))sin(pi/12)` ===> (7)
This a binomial with `bar(AC)_1 = 7.12016, bar(AC)_2=−7.12016)`
Since we are calculate a "length" the positive will do, `bar(AC)_1 = 7.12016` Now we also use (1) to calculate `bar(AB)`

`bar(AB) = 1/sin(pi/12)6/7.12016 = 3.25586` ===> (8)

`bar(BC) = 1/sin(8pi/12)6/7.12016 = 0.97304` ===> (9)

Now use the Triangle Area Formula with r:

`A_Delta=3= 1/2 r (bar(AC)+bar(AB)+bar(BC))=` solve for r
`r = 6/((bar(AC)+bar(AB)+bar(BC))` ===> (10)
`r = 6/((7.12016 )+(3.25586)+(0.97304))= 0.52868`
Now calculate the Area of the circle:
`A_(@) = pir^2 = pi(0.52868)^2 = .2795pi`