# A triangle has sides A, B, and C. The angle between sides A and B is (2pi)/3 and the angle between sides B and C is pi/12. If side B has a length of 16, what is the area of the triangle?

##### 1 Answer
Mar 5, 2016

Two angles of the triangle are$\left(2 \frac{\pi}{3} , \frac{\pi}{12}\right)$
So the third angle between sides A and C $= \pi - 2 \frac{\pi}{3} - \frac{\pi}{12} = \frac{12 \pi - 8 \pi - \pi}{12} = 3 \cdot \frac{\pi}{12} = \frac{\pi}{4}$
Pl consider the fig. below

From Properties of triangle we know the sides of a triangle are proportional to the $\sin e$ of opposite angle
$\therefore$$\frac{A}{\sin} \left(\frac{\pi}{12}\right) = \frac{16}{\sin} \left(\frac{\pi}{4}\right)$
$\implies A = 16 \sin \frac{\frac{\pi}{12}}{\sin} \left(\frac{\pi}{4}\right) = 16 \sqrt{2} \sin \left(\frac{\pi}{12}\right)$

Now area of the triangle $= \left(\frac{1}{2}\right) \cdot A \cdot B \sin \left(2 \frac{\pi}{3}\right)$
$= \left(\frac{1}{2}\right) \cdot 16 \sqrt{2} \sin \left(\frac{\pi}{12}\right) \cdot 16 \cdot \sin \left(2 \frac{\pi}{3}\right)$
$= 128 \sqrt{2} \frac{\sqrt{3}}{2} \cdot \sin \left(\frac{\pi}{12}\right)$
$= 64 \sqrt{6} \cdot \sin \left(\frac{\pi}{12}\right) = 40.6$squnit