A triangle has sides A,B, and C. If the angle between sides A and B is `(5pi)/12`, the angle between sides B and C is `pi/6`, and the length of B is 15, what is the area of the triangle?

The angle between `A` and `B` is

`=pi-(5/12pi+2/12pi)`

`=pi-(7/12pi)`

`=5/12pi`

So, the triangle is isoceles

The side `B` is `=15`

Therefore, the side `C` is `=15`

`sin(1/6pi)=1/2`

The area of the triangle is

`A=1/2*15*15sin(1/6pi)`

`=225/4=56.25`

Here's all the info we know (I found that last angle by subtracting all the other angles from `2pi)`:

So, here's what we need to know for the area:

Once we find these values, we'll be able to use the formula `area=1/2(b xx h)`

Let's work on finding the height, `h`.

To do that, we just need to use `sin(pi/6)=h/15`, or `7.5=h`

Now we know the height, all that's left is to find the base, or `c`

First, let's put all our info in a table:

length `color(white)(0000)` angle
`A = ? color(white)(0000) A= pi/6`
`B = 15 color(white)(0000) B = (17pi)/12`
`C = ? color(white)(0000) A= (5pi)/12`

Let's use law of sines

`(sin((5pi)/12))/(C) = (sin((17pi)/12))/(15)`

`c = -15`, but because this problem is dealing in distances, we cannot have a negative length, so `c=15`

Now let's use our formula: `area=1/2(b xx h)`

`area = (15 xx 7.5)/2`

`area = 56.25` units squared