# A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/6 and the angle between sides B and C is pi/12. If side B has a length of 21, what is the area of the triangle?

$\angle B = \pi - \left(\frac{5 \pi}{6} + \frac{\pi}{12}\right) = \frac{\pi}{12} , \frac{a}{\sin} \left(\frac{\pi}{12}\right) = \frac{21}{\sin} \left(\frac{\pi}{12}\right) \to a = \frac{21}{\sin} \left(\frac{\pi}{12}\right) \cdot \sin \left(\frac{\pi}{12}\right) = 21$
$A r e a = \frac{1}{2} a b \sin C = \frac{1}{2} \left(21\right) \left(21\right) \sin \left(\frac{5 \pi}{6}\right) = 110.25 u n i {t}^{2}$
First, add the two given angles and subtract from $\pi$ to get the third angle. Then find one of the other two sides. In this case I found side a by using the sine rule. So to find the area of the triangle I use side a and b and angle C and substitute in to the angle formula then calculate.