# A triangle has sides A, B, and C. The angle between sides A and B is (pi)/2 and the angle between sides B and C is pi/12. If side B has a length of 17, what is the area of the triangle?

$\angle B = \pi - \left(\frac{\pi}{2} + \frac{\pi}{12}\right) = \frac{5 \pi}{12} , \to \frac{a}{\sin} A = \frac{b}{\sin} B \to \frac{a}{\sin} \left(\frac{\pi}{12}\right) = \frac{17}{\sin} \left(\frac{5 \pi}{12}\right) \to a = \frac{17 \sin \left(\frac{\pi}{12}\right)}{\sin} \left(\frac{5 \pi}{12}\right) = 4.555 \ldots$
$A r e a \triangle = \frac{1}{2} \left(a\right) \left(b\right) \sin C = \frac{1}{2} \left(4.555 \ldots\right) \left(17\right) \sin \left(\frac{\pi}{2}\right) \approx 38.72$
First subtract the given angles from $\pi$ to find the third angle. In order to find the area of the triangle we need to find the measurement of one of the other two sides. So I use the law of sines to find side a and then put it into the area formula and calculate. Note I use angle C since sides a and b are in my calculation.