# A triangle has sides A, B, and C. Sides A and B have lengths of 6 and 14, respectively. The angle between A and C is (17pi)/24 and the angle between B and C is  (pi)24. What is the area of the triangle?

$\frac{1}{2} \cdot 6 \cdot 14 \cdot \sin \left(\frac{\pi}{4}\right) = 21 \sqrt[2]{2}$
The angle between A and B is given by $\theta = \pi - 17 \frac{\pi}{24} - \frac{\pi}{24} = \frac{24 \pi - 18 \pi}{24} = \frac{\pi}{4}$
The area is given by $\frac{1}{2} \cdot A \cdot B \cdot \sin \left(\theta\right)$