# A triangle has sides A, B, and C. The angle between sides A and B is (3pi)/4 and the angle between sides B and C is pi/12. If side B has a length of 27, what is the area of the triangle?

Jul 15, 2018

$A = \frac{1}{4} \cdot {27}^{2} \cdot \sqrt{2} \cdot \left(\sqrt{3} + 1\right)$

#### Explanation:

At first we will compute the third angle:
$\pi - \frac{3}{4} \cdot \pi - \frac{\pi}{12} = \frac{12 \pi - 9 \pi - \pi}{12} = \frac{\pi}{6}$
with the Theorem of sines weget

$a = \frac{27 \cdot \sin \left(\frac{\pi}{6}\right)}{\sin \left(\frac{\pi}{12}\right)}$
note that

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

$\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$
Now we use the Formula
$A = \frac{1}{2} \cdot a \cdot b \cdot \sin \left(\gamma\right)$

so we get

$A = \frac{1}{2} \cdot {27}^{2} \cdot \frac{\sqrt{2}}{\sqrt{3} - 1}$
this is

$A = \frac{1}{4} \cdot {27}^{2} \cdot \left(\sqrt{3} + 1\right)$

We have used that
$\frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{2}$