# A triangle has sides A, B, and C. The angle between sides A and B is pi/3 and the angle between sides B and C is pi/12. If side B has a length of 1, what is the area of the triangle?

Jan 19, 2016

$A \approx 0.116$

#### Explanation: The area of a triangle $A = \frac{1}{2} \cdot B a s e \cdot h e i g h t$
In this case the base $B = 1$
If we name the two portions of the base on either side of the perpendicular $h$ as $x$ and $y$, then $B = x + y = 1$

We then use the trigonometric relations
$\frac{h}{x} = \tan \left(\frac{\pi}{12}\right)$
$\frac{h}{y} = \tan \left(\frac{\pi}{3}\right)$

We now have three equations and three unknowns so we can solve to find $h$
$x = \frac{h}{\tan} \left(\frac{\pi}{12}\right)$
$y = \frac{h}{\tan} \left(\frac{\pi}{3}\right)$
$\therefore \frac{h}{\tan} \left(\frac{\pi}{12}\right) + \frac{h}{\tan} \left(\frac{\pi}{3}\right) = 1$

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

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

$A = \frac{1}{2} \cdot 1 \cdot \frac{\tan \left(\frac{\pi}{12}\right) \tan \left(\frac{\pi}{3}\right)}{\tan \left(\frac{\pi}{3}\right) + \tan \left(\frac{\pi}{12}\right)}$

$A \approx \frac{0.268 \cdot 1.732}{2 \left(1.732 + 0.268\right)}$

$A \approx \frac{0.464}{4} \approx 0.116$