# 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 13, what is the area of the triangle?

Area of the triangle is $30.93$ Sq units
The angle between sides B & C is $\angle A = \frac{\pi}{12} = \frac{180}{12} = {15}^{0}$
The angle between sides A & B is $\angle C = 3 \cdot \frac{\pi}{4} = 3 \cdot \frac{180}{4} = {135}^{0}$
So Angle $\angle B = 180 - \left(135 + 15\right) = {30}^{0}$ Using sine law we get $\frac{B}{\sin} B = \frac{A}{\sin} A \mathmr{and} \frac{13}{\sin} 30 = \frac{A}{\sin} 15 \mathmr{and} A = 13 \cdot \sin \frac{15}{\sin} 30 = 6.73$ Area $= \frac{A . B . \sin C}{2} = \frac{13 \cdot 6.73 \cdot \sin 135}{2} = 30.93$ sq.units