A triangle has sides A,B, and C. If the angle between sides A and B is `(pi)/12`, the angle between sides B and C is `pi/6`, and the length of B is 5, what is the area of the triangle?

Given: A triangle with angles `pi/12 (15^o), pi/6 (30^o), (3 pi)/4 (135^o)` and `B=5`

There are multiple ways to solve this problem. You can use the formula: `Area = 1/2 "Base" * "height"` or you can use Heron's formula which requires you to find the length of all sides.

Find the area using `Area = 1/2 "Base" * "height"`:

Find the height:

`" "sin (pi/12) = h/5 " so "h = 5 sin( pi/12)~~1.294`

Find the base A:

You can use the Pythagorean Theorem or trigonometry to find `A + h`:

`sin (30^o + 45^o) = sin(75^o) = (A+h)/5`

`A +h = 5 sin 75^o`

`A = 5 sin 75^o - h = 5 sin 75^o - 5 sin( pi/12) = 5(sin 75^o -sin(pi/12)) ~~ 3.536`

`Area = 1/2 * 3.536 * 1.294~~ 2.29 "units"^2`

Find the Area using Heron's Formula:

`s = (A + B + C)/2; Area = sqrt(s(s-A)(s-B)(s-C))`

Since we have a `45^o - 45^o - 90^o` triangle formed with the height, we have the ratio: `h: h: h sqrt(2)` for side lengths.

This means `C = h sqrt(2) = 5 sqrt(2) sin (pi/12) ~~1.83`

From above, `A = 5(sin 75^o -sin(pi/12))~~3.536`

`s ~~ 5.183`

`Area = sqrt(5.183(1.647)(.183)(3.353)) ~~2.29 "units"^2`