# A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 4, what is the area of the triangle?

Feb 27, 2016

pl,see below

#### Explanation:

The angle between sides A and B $= 5 \frac{\pi}{12}$
The angle between sides C and B $= \frac{\pi}{12}$
The angle between sides C and A $= \pi - 5 \frac{\pi}{12} - \frac{\pi}{12} = \frac{\pi}{2}$
hence the triangle is right angled one and B is its hypotenuse.
Therefore side A = $B \sin \left(\frac{\pi}{12}\right) = 4 \sin \left(\frac{\pi}{12}\right)$
side C = $B \cos \left(\frac{\pi}{12}\right) = 4 \cos \left(\frac{\pi}{12}\right)$
So area$= \frac{1}{2} A C \sin \left(\frac{\pi}{2}\right) = \frac{1}{2} \cdot 4 \sin \left(\frac{\pi}{12}\right) \cdot 4 \cos \left(\frac{\pi}{12}\right)$
$= 4 \cdot 2 \sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{12}\right)$
$= 4 \cdot \sin \left(2 \frac{\pi}{12}\right)$
$= 4 \cdot \sin \left(\frac{\pi}{6}\right)$
$= 4 \cdot \frac{1}{2}$ =2 sq unit