# A triangle has sides A, B, and C. Sides A and B have lengths of 9 and 3, respectively. The angle between A and C is (13pi)/24 and the angle between B and C is  (pi)/8. What is the area of the triangle?

use the formula area = $\frac{1}{2} A . B \sin \theta$
here the angle between sides Aand B = $\theta$
So, $\theta = \pi - 13 \frac{\pi}{24} - \frac{\pi}{8} = \pi - 16 \frac{\pi}{24} = \frac{\pi}{3}$
Hence area = $\frac{1}{2} \cdot 9 \cdot 3 \cdot \sin \left(\frac{\pi}{3}\right)$= 6.75 squnit