# A triangle has sides A, B, and C. The angle between sides A and B is pi/6 and the angle between sides B and C is pi/12. If side B has a length of 9, what is the area of the triangle?

Jan 27, 2016

A = 1/2B*B*(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)

#### Explanation:

The area of a triangle is given by $A = \frac{1}{2} B h$ where $B$ is the base and $h$ is the perpendicular height.

We also know that the sum of the angles in a triangle is ${180}^{o} = \pi$

The angle between $B$ and $C$ is therefore $\left(\pi - \frac{\pi}{12} - \frac{\pi}{6}\right)$
=(12pi-pi-2pi)/12) = 9pi/12 =(3pi)/4

If $h$ intersects with $B$ so that $x + y = B$ then
$\frac{h}{B - y} = \tan \left(\frac{\pi}{12}\right)$ and $\frac{h}{y} = \tan \left(\frac{3}{4} \pi\right)$

Then $y = \frac{h}{\tan} \left(\frac{3}{4} \pi\right)$
$\therefore h = \left(B - \frac{h}{\tan} \left(\frac{3}{4} \pi\right)\right) \tan \left(\frac{\pi}{12}\right)$

$h = B \tan \left(\frac{\pi}{12}\right) - h \tan \frac{\frac{\pi}{12}}{\tan} \left(\frac{3}{4} \pi\right)$

So $h \left(1 + \tan \frac{\frac{\pi}{12}}{\tan} \left(\frac{3}{4} \pi\right)\right) = B \tan \left(\frac{\pi}{12}\right)$

$h \left(\frac{\tan \left(\frac{3}{4} \pi\right) + \tan \left(\frac{\pi}{12}\right)}{\tan} \left(\frac{3}{4} \pi\right)\right) = B \tan \left(\frac{\pi}{12}\right)$

:. h = B(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)

So the area A = 1/2B*B(tan(pi/12)tan(3/4pi))/(tan(3/4pi) + tan(pi/12)