Question

A triangle has sides A, B, and C. The angle between sides A and B is `pi/4` and the angle between sides B and C is `pi/12`. If side B has a length of 17, what is the area of the triangle?

Answer 1

I get `{289(3-\sqrt{3})}/12`, about `30.5` square units.

Explanation:

First find the third angle, between sides A and C:

`\pi/4+\pi/12+\theta=\pi`

`\theta={2\pi}/3`

`\sin\theta={\sqrt{3}}/2` for the next step

Now, Use the Law of Sines to find a second side. We choose side A and must know the sine of `\pi/12`:

`\sin(\pi/12)=\sin(\pi/4-\pi/6)`

`=\sin(\pi/4)\cos(\pi/6)-\cos(\pi/4)\sin(\pi/6)`

`=(\sqrt{2}/2)((\sqrt{3}/2)-(1/2)(\sqrt{2}/2)`

`={\sqrt{6}-\sqrt{2}}/4`

And then the Law of Sines gives:

`A/{({\sqrt{6}-\sqrt{2}}/4)} ={17}/{({\sqrt{3}}/2}`

`A={17(3\sqrt{2}-\sqrt{6})}/6`

Now that we have two sides A and B, we can take half their product time the sine of the included angle:

Area=`(1/2)(17)({17(3\sqrt{2}-\sqrt{6})}/6)({\sqrt{2}}/2)`

`={289(3-\sqrt{3})}/12`.