A triangle has sides A, B, and C. The angle between sides A and B is `(7pi)/12`. If side C has a length of `1` and the angle between sides B and C is `pi/12`, what is the length of side A?

You can use the law of sines.

The angle between sides `A` and `B` is the angle opposing the side `C`. Let's call this angle `gamma`.

Similarly, the angle between sides `B` and `C` is the one opposing the side `A`. Let's call this angle `alpha`.

Even though we don't need it for this particular task, let's also call the last remaining angle `beta`, this one is opposing the side `B` .

According to the law of sines, the following relation between the sides and the opposite angles exists:

`A / sin alpha = B / sin beta = C / sin gamma`

In our case, we only need to look at `A`, `C`, `alpha` and `gamma`:

`A / sin alpha = C / sin gamma`

`A / sin (pi / 12) = 1 / sin ((7 pi)/12)`

`<=> A = sin (pi / 12) / sin ((7pi) / 12)`

So, the only thing left to do is compute `sin (pi/12)` and `sin((7pi)/12)`.

Let me show you how to do this without the calculator but with some basic knowledge about `sin` and `cos` and using the identity

`sin (a-b) = sin(a)cos(b) - cos(a)sin(b)`.

You need to express `pi/12` as a sum or difference of simpler values:

`sin (pi/12) = sin (pi/3 - pi/4)`

`= sin(pi/3) cos(pi/4) - cos(pi/3)sin(pi/4)`

`= sqrt(3)/2 * 1/sqrt(2) - 1 / 2 * 1 / sqrt(2)`

`= 1/(2 sqrt(2)) (sqrt(3) - 1)`

Similarly,

`sin((7 pi)/12) = sin(pi/3 + pi/4)`

`= sin(pi/3) cos(pi/4) + cos(pi/3) sin(pi/4)`

`= sqrt(3) / 2 * sqrt(2) / 2 + 1 / 2 * sqrt(2) / 2`

`= 1/(2 sqrt(2)) (sqrt(3) + 1)`

Thus, the length of the side `A` is

`A = sin (pi / 12) / sin ((7pi) / 12)`

`= (1/(2 sqrt(2)) (sqrt(3) - 1) )/(1/(2 sqrt(2)) (sqrt(3) + 1))`

`= (sqrt(3) - 1) / (sqrt(3) + 1)`

`= ((sqrt(3) - 1) color(blue)((sqrt(3)-1))) / ((sqrt(3) + 1) color(blue)((sqrt(3)-1)))`

`= (sqrt(3) -1)^2 / ((sqrt(3))^2 - 1^2 )`

`= (3 - 2sqrt(3) + 1) / 2`

`= 2 - sqrt(3)`