# What is the physical meaning of $\alpha$ in this disintegration problem?

I am taking a course where I have to solve the following problem related to the disintegration of a baryon in a proton and a pion $(\Lambda \to p^+ + \pi^-)$:

Assume that a $\Lambda$ particle with energy $10 ~GeV$ is produced. What mean distance will it travel before decaying? What is the range of possible values for the angle $\theta$ (where $\theta$ is the initial angle between the trajectories of the $p^+$ and $\pi^-$ particles) is the decaying is isotropic in the proper system?

As a data I have the masses of every particle and the half life of the baryon.

I have the solution of my professor to the problem but there is something that I don't understand when he's trying to calculate the range of values for $\theta$. He considers the system of the centre of inertia of and draws the following diagram:

This is what I don't understand. I understand the calculations that follow but I don't understand what is the meaning of the angle $\alpha$. What are the possible values that $\alpha$ can take? What is its physical meaning?

Since I can't comment, I can't ask for specification. Most probably problem is following. In the lab system $\Lambda$ with total energy $E= 10~GeV$ flies along x axis and after some time decays into proton and pion. In the $\Lambda$ rest frame, which is oriented the same way (x, y, z coordinates), those daughter particles must fly back to back isotropically, so any direction is allowed. So there is only one parameter describing its decay, which is $\alpha$ angle in respect to one of the axes. Most convenient is to choose the x-axis as well. Range of the $\alpha$ angle is $0$ to $2\pi$.
Now, when you move from $\Lambda$ rest frame to lab frame, you must boost your daughter particles by $\gamma = E/m$. Therefore the back-to-back geometry in the $\Lambda$ rest fame will turn into cone with angle $\theta$ which will be a function of $\alpha$.