Answer:
[tex]$L=\frac{v^2}{2-2\cos(\theta)}$[/tex]
Step-by-step explanation:
Solve for [tex]L[/tex] in
[tex]$L-L \cos(\theta)=\frac{v^2}{2} \Longrightarrow L(1-\cos(\theta))=\frac{v^2}{2}$[/tex]
[tex]$L=\frac{v^2}{2(1-\cos(\theta))}=\frac{v^2}{2-2\cos(\theta)}$[/tex]
