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

Jun 25, 2018

$A = \frac{72 \cdot \sqrt{3}}{1 + \sqrt{3}}$

#### Explanation:

We use the Formula

$A = \frac{1}{2} \cdot a \cdot b \cdot \sin \left(\gamma\right)$
where $b , \gamma$ is given. So we must compute the side length of $a$:

The third angle is given by $5 \frac{\pi}{12}$
so we can use the Theorem of sines:

$\sin \frac{\frac{5}{12} \cdot \pi}{\sin} \left(\frac{\pi}{3}\right) = \frac{12}{a}$

From here we get

$a = 12 \sin \frac{\frac{\pi}{3}}{\sin} \left(5 \cdot \frac{\pi}{12}\right)$
putting Things together we get

$A = \frac{1}{2} \cdot 12 \cdot 12 \cdot \sin \left(\frac{\pi}{3}\right) \cdot \sin \frac{\frac{\pi}{4}}{\sin} \left(5 \cdot \frac{\pi}{12}\right)$

Note that

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

$\sin \left(5 \cdot \frac{\pi}{12}\right) = \frac{1 + \sqrt{3}}{2 \cdot \sqrt{2}}$

$\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
thus

$A = \frac{72 \cdot \sqrt{3}}{1 + \sqrt{3}}$