= Solution
Nonrelativistic <momentum conservation> gives $\mathbf u=\mathbf v_1+\mathbf v_2$, while an <elastic collision> conserves <kinetic energy>, $u^2=v_1^2+v_2^2$. Squaring the <momentum> equation yields $\mathbf v_1\cdot\mathbf v_2=0$. \b[The two nonzero outgoing velocities are perpendicular.] In the degenerate case where one outgoing particle is at rest, its direction is undefined.
For relativistic collisions conserve total <four-momentum>, with $E=\gamma mc^2$, $\mathbf p=\gamma m\mathbf v$ and $E^2-p^2c^2=m^2c^4$. Rest energies must be included; for <photons> $E=pc$.
For the <symmetric relativistic equal-mass scattering angle>, the incident <momentum> is $mc\sqrt{\alpha(\alpha+2)}$. Energy conservation gives each final particle $E_f=(1+\alpha/2)mc^2$, hence $p_f=(mc/2)\sqrt{\alpha(\alpha+4)}$. Transverse momenta cancel and $2p_f\cos\theta=p_{\rm in}$, giving
$$
\cos\theta=\sqrt{\frac{\alpha+2}{\alpha+4}},\qquad\boxed{\tan\theta=\sqrt{\frac2{\alpha+2}}.}
$$
The nonrelativistic limit has $\theta\to45^\circ$, so the outgoing angle is $90^\circ$ as above.
For the <symmetric two-photon annihilation angle> in the specified <electron-positron annihilation>, each <photon> has energy $E_\gamma=(\alpha+2)m_ec^2/2$. <Momentum conservation> gives $2(E_\gamma/c)\cos\phi=m_ec\sqrt{\alpha(\alpha+2)}$, so
$$
\cos\phi=\sqrt{\frac\alpha{\alpha+2}},\qquad\boxed{\tan\phi=\sqrt{\frac2\alpha}\quad(\alpha>0).}
$$
As $\alpha\to0^+$ the two <photons> become back-to-back, transverse to the limiting incident direction; as $\alpha\to\infty$ they are forward collimated. Equal <photon> energies follow from the specified equal angles, and are not a general property of annihilation in a moving laboratory frame.
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