Computing vector to equal vector forces/magnitudes I have this problem with its respective solution:

Assuming that the superior vector is $\overrightarrow{A}$, the middle vector is $\overrightarrow{B}$, and the inferior vector is $\overrightarrow{F}$
My computation was:
$$\overrightarrow{A} \cdot \overrightarrow{B}=\overrightarrow{B} \cdot \overrightarrow{F}\\
|\overrightarrow{A}||\overrightarrow{B}|cos(40)=|\overrightarrow{B}||\overrightarrow{F}|cos(35)\\
|\overrightarrow{F}| = \frac{20000*cos(40)}{cos(35)} \space\space\space\space \text{(given} |\overrightarrow{A}|=20000 \text{)} \\
|\overrightarrow{F}| = 18703
$$
Where am I wrong? I'm looking for a vector solution.

Although I don't recommend it for this application, one can get a vector solution using $$\vec{A}\times \vec{B}=S\vec{F}\times \vec{B}$$
Where variable $S$ is the unknown force.
$$
\vec{A} \, =  \, \left( \begin{align}20000 \; \cos \left( 2 \cdot \frac{\pi }{9} \right) \\ 20000 \; \sin \left( 2 \cdot \frac{\pi }{9} \right) \\ 0 \end{align} \right)$$
$$
\vec{B} \, =  \, \left( \begin{align}1 \\ 0 \\ 0 \end{align} \right)
$$
$$
\vec{F} \, =  \, \left( \begin{align}S \; \operatorname{cos} \left( 7 \cdot \frac{\pi }{36} \right) \\ S \; \operatorname{sin} \left( 7 \cdot \frac{\pi }{36} \right) \\ 0 \end{align} \right)
$$

Now $$\vec{A}\times\vec{B}=\left( \begin{align}0 \\ 0 \\ -20000 \; \sin \left( \frac{2}{9} \; \pi  \right) \end{align} \right)
$$
$$\vec{F}\times\vec{B}=\left( \begin{align}0 \\ 0 \\ -S \; \sin \left( \frac{7}{36} \; \pi  \right) \end{align} \right)$$
Solving for $S$
$$ \left\{ S = 20000 \cdot \frac{\sin \left( 2 \cdot \frac{\pi }{9} \right)}{\sin \left( 7 \cdot \frac{\pi }{36} \right)} \right\}=22413.32 $$
Just because you can, doesn't mean you should.
