Question

A triangle has sides A, B, and C. The angle between sides A and B is `pi/6` and the angle between sides B and C is `pi/12`. If side B has a length of 9, what is the area of the triangle?

Answer 1

`A = 1/2B*B*(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)`

Explanation:

The area of a triangle is given by `A=1/2Bh` where `B` is the base and `h` is the perpendicular height.

We also know that the sum of the angles in a triangle is `180^o = pi`

The angle between `B` and `C` is therefore `(pi - pi/12 - pi/6)`
`=(12pi-pi-2pi)/12) = 9pi/12 =(3pi)/4`

If `h` intersects with `B` so that `x +y = B` then
`h/(B-y) = tan(pi/12)` and `h/y = tan(3/4pi)`

Then `y=h/tan(3/4pi)`
`:. h = (B-h/tan(3/4pi))tan(pi/12)`

`h = Btan(pi/12) - htan(pi/12)/tan(3/4pi)`

So `h(1+tan(pi/12)/tan(3/4pi)) = Btan(pi/12)`

`h((tan(3/4pi) + tan(pi/12))/tan(3/4pi)) = Btan(pi/12)`

`:. h = B(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)`

So the area `A = 1/2B*B(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)`