I need to find the value on the $v_{1}$ side of the triangle, the book's answer is $v_{1} = V cos(\psi) + v'_{1} cos (\theta - \psi)$, but I couldn't understand how you get to that result. I used the cosine law but it didn't work, how can I solve it?
The main exercise is about elastic collisions, in which I need to relate the initial kinetic energy and the final kinetic energy of the particle in the LAB framework :
$$ \frac{T_{1}}{T_{0}}=\frac{m^2}{(m_{1}+m_{2})^2}.S^2 $$
$$ S= \cos(\psi)+\frac{\cos(\theta-\psi)}{(m_{1}/m_{2})} $$
thereby
$$ T_{0}=\frac{m_{1}u_{1}^2}{2} $$
$$ T_{1}=\frac{m_{1}v_{1}^2}{2} $$
$$ \frac{T_{1}}{T_{0}}=\frac{v_{1}^2}{u_{1}^2} $$
and the book says that by the triangle diagram you get the relationship $v_{1} = V cos(\psi) + v'_{1} cos (\theta - \psi)$
The figure is the final state of mass $m_{1}$ for the elastic collision of two particles for the case $V<v'_{1}$ for which there is one trajectory.
$V$ is the velocity of the center of mass in the LAB framework.
That is question 9.38 from the book Classical Dynamics of Particles and Systems by Marion, Thornton.
Taking inspiration from Piquito, in order to see the answer clearly, draw the perpendicular from the endpoint of $V$ (where $\theta$ is) to $v_1$.
That divides $v1$ into 2 segments: The first is equal to $V\cos(\psi)$, and the second is equal to $v_1'\cos(\theta-\psi)$. These two segments add up to the expression in your book's answer: $v1=V\cos(\psi)+v_1'\cos(\theta-\psi)$.