= Solution
In the pion rest frame, <conservation of momentum> makes the muon and massless neutrino momenta equal and opposite, of magnitude $p$. The neutrino energy is $pc$, while the muon satisfies $E^2=p^2c^2+m^2c^4$. Energy conservation gives $E+pc=Mc^2$. Squaring after substituting $E=Mc^2-pc$ yields
$$
p=\frac{M^2-m^2}{2M}c,\qquad E=\frac{M^2+m^2}{2M}c^2.
$$
The <relativistic two-body decay> speed is thus
$$
\boxed{u=\frac{pc^2}{E}=c\frac{M^2-m^2}{M^2+m^2}}.
$$
Choose $S'$ moving with velocity $+v$ along $x$ relative to $S$, so the pion has velocity $-v$ in $S'$ and speed $v$. With $\Gamma_v=(1-v^2/c^2)^{-1/2}$, the four-dimensional <Lorentz transformation> is
$$
\begin{pmatrix}ct'\\x'\\y'\\z'\end{pmatrix}
=\begin{pmatrix}\Gamma_v&-\Gamma_vv/c&0&0\\-\Gamma_vv/c&\Gamma_v&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}
\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}.
$$
The muon's <four-momentum> is $(E/c,0,p,0)$ before the boost. The same <matrix> gives
$$
E'=\Gamma_vE,\qquad
\boxed{\boldsymbol p'=\left(-\Gamma_v\frac{vE}{c^2},\ p,\ 0\right)}.
$$
Velocity and momentum have the same spatial direction. The unsigned deflection angle from the positive $y$ axis therefore satisfies
$$
\boxed{\tan\theta=\frac{|p'_x|}{p'_y}
=\frac{M^2+m^2}{M^2-m^2}\frac{v}{\sqrt{c^2-v^2}}}.
$$
Choosing the opposite boost makes the pion move along positive $x$ and reverses $p'_x$, leaving this speed-based angle formula unchanged.
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