= Solution
In the <laboratory frame> the incoming particle and the composite move along $+x$, while the target is initially at rest. In the <equal-energy two-photon decay of a moving particle>, zero total transverse <momentum> makes the two <photon> directions symmetric about $+x$. Each <photon> has <energy> $2mc^2$ and <momentum> magnitude $2mc$, so longitudinal <momentum conservation> gives $2(2mc)\cos\theta=2.4mc$. Therefore
$$
\boxed{\theta=\arccos0.6=53.130^\circ,\qquad\text{photon opening angle}=2\theta=106.260^\circ.}
$$
Their momenta can be written $(1.2,\pm1.6)mc$. The opening <angle> is not $180^\circ$ in this frame, because the parent carried nonzero total <momentum>.
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