# 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?

Apr 1, 2018

I get $\frac{289 \left(3 - \setminus \sqrt{3}\right)}{12}$, about $30.5$ square units.

#### Explanation:

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

$\setminus \frac{\pi}{4} + \setminus \frac{\pi}{12} + \setminus \theta = \setminus \pi$

$\setminus \theta = \frac{2 \setminus \pi}{3}$

$\setminus \sin \setminus \theta = \frac{\setminus \sqrt{3}}{2}$ 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 $\setminus \frac{\pi}{12}$:

$\setminus \sin \left(\setminus \frac{\pi}{12}\right) = \setminus \sin \left(\setminus \frac{\pi}{4} - \setminus \frac{\pi}{6}\right)$

$= \setminus \sin \left(\setminus \frac{\pi}{4}\right) \setminus \cos \left(\setminus \frac{\pi}{6}\right) - \setminus \cos \left(\setminus \frac{\pi}{4}\right) \setminus \sin \left(\setminus \frac{\pi}{6}\right)$

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

$= \frac{\setminus \sqrt{6} - \setminus \sqrt{2}}{4}$

And then the Law of Sines gives:

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

$A = \frac{17 \left(3 \setminus \sqrt{2} - \setminus \sqrt{6}\right)}{6}$

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

Area=$\left(\frac{1}{2}\right) \left(17\right) \left(\frac{17 \left(3 \setminus \sqrt{2} - \setminus \sqrt{6}\right)}{6}\right) \left(\frac{\setminus \sqrt{2}}{2}\right)$

$= \frac{289 \left(3 - \setminus \sqrt{3}\right)}{12}$.