<s>[INST] Triangle of forces Forces equal to $5P$, $12P$ and $13P$ acting on a particle are in equilibrium ;find ,by geometric construction and by calculation ,the angles between their directions?
I have an problem that, With three forces how can I use Lami's rule?
 [/INST] let $\vec x\parallel\vec F_1, \vec x \perp \vec y$
$\vec x:$
$\angle \vec F_2=\alpha$, $\angle \vec F_3=\beta$
$\vec y:$
$\angle \vec F_2=\frac {\pi}{2} - \alpha$, $\angle \vec F_3=\frac {\pi}{2} - \beta$
$\begin{equation*}
 \begin{cases}
   F_1 + F_2 \cdot \cos \alpha + F_3 \cdot \cos \beta = 0
\\
0 + F_2 \cdot \sin \alpha + F_3 \cdot \sin \beta = 0
 \end{cases}
\end{equation*}$
$\begin{equation*}
 \begin{cases} 
   F_1 + F_2 \cdot \cos \alpha + F_3 \cdot \cos \beta = 0
\\
12 \cdot \sin \alpha = -13 \cdot  \sin \beta
 \end{cases}
\end{equation*}$
$\sin \alpha = - \frac {13}{12} \cdot \sin \beta$
$\cos^2 \alpha = 1 - \frac {169}{144} \cdot \sin^2 \beta=\frac {169}{144} \cdot \cos^2 \beta - \frac {25}{144}$
$ 5 + 12 \cdot \cos \alpha + 13 \cdot \cos \beta = 0$
$ 12 \cdot \cos \alpha =- 13 \cdot \cos \beta - 5$
$ 144 \cdot \cos^2 \alpha =(13 \cdot \cos \beta + 5)^2$
$169 \cdot \cos^2 \beta - 25 = 169 \cdot \cos^2 \beta + 130 \cdot \cos \beta + 25$ 
$130 \cdot \cos \beta = -50$ 
$\cos \beta = \frac {5}{13}, \beta < \pi$
$ \beta = 1.176$
$ \sin \alpha = -1$
$\alpha = \frac {3 \cdot \pi}{2}$
Answer:
$\angle \vec F_1 \vec F_2 = \frac {3 \cdot \pi}{2}, \angle \vec F_1 \vec F_3 = 1.176$
EDIT: Stupid mistake
</s>