# 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 7, what is the area of the triangle?

Feb 4, 2016

$S = 49 \left(1 - \frac{\sqrt{3}}{2}\right) \cong 6.5648$

#### Explanation:

It is a right triangle with catheti $A$ and $B$ and hypotenuse $C$.
Knowing the length of $B = 7$ and an angle between $B$ and $C$ equaled $\frac{\pi}{12}$, we can calculate $A$ as follows.

Since $\frac{A}{B} = \tan \left(\frac{\pi}{12}\right)$,
it follows that $A = B \cdot \tan \left(\frac{\pi}{12}\right)$

Area of a triangle is
$S = \frac{1}{2} \left(A \cdot B\right) = \frac{1}{2} {B}^{2} \tan \left(\frac{\pi}{12}\right)$

The latter can be simplified using a formula
$\tan \left(\phi\right) = \frac{1 - \cos \left(2 \phi\right)}{\sin} \left(2 \phi\right)$
from which follows:
$\tan \left(\frac{\pi}{12}\right) = \frac{1 - \cos \left(\frac{\pi}{6}\right)}{\sin} \left(\frac{\pi}{6}\right) = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}$

Therefore, the area of a triangle is
$S = \left(\frac{1}{2}\right) {7}^{2} \left(2 - \sqrt{3}\right) = 49 \left(1 - \frac{\sqrt{3}}{2}\right)$