= Solution
For two <photons>, <four-momentum conservation> and their vanishing rest <masses> give the <diphoton invariant mass>
$$
M^2=2E_1E_2(1-\cos\alpha).
$$
Taking a logarithmic differential,
$$
2\frac{dM}{M}=\frac{dE_1}{E_1}+\frac{dE_2}{E_2}+\frac{\sin\alpha}{1-\cos\alpha}\,d\alpha.
$$
For small mutually independent measurement errors, <propagation of uncertainty> consequently gives
$$
\boxed{\left(\frac{\sigma_M}{M}\right)^2=\frac14\left[\left(\frac{\sigma_{E_1}}{E_1}\right)^2+\left(\frac{\sigma_{E_2}}{E_2}\right)^2+\left(\frac{\sin\alpha\,\sigma_\alpha}{1-\cos\alpha}\right)^2\right].}
$$
\b[Printed-formula correction.] The source's left-hand side, if it means $(\Delta M/M)^2$ as its description suggests, is missing the factor $1/4$ on the right. Its right-hand side instead equals $\operatorname{Var}(M^2)/M^4$ to first order. A concrete counterexample is a fixed opening angle and an exactly known $E_2$: because $M\propto\sqrt{E_1}$, a small $1\%$ fractional error in $E_1$ produces a $0.5\%$ fractional error in $M$, not $1\%$. Likewise, interpreting $\Delta M^2$ literally as a small change in $M^2$ would make the unsquared left-hand side incompatible with the squared terms. Correlated errors require the covariance terms as well.
The space angle is determined by the two polar angles and azimuthal separation:
$$
\cos\alpha=\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\cos(\Delta\phi).
$$
At $\theta_1=\theta_2=45^\circ$ and $\Delta\phi=180^\circ$, $\cos\alpha=0$ and $\alpha=90^\circ$. The angular coefficient is therefore one. Equal contribution of opening-angle uncertainty and the \i[combined] two independent $1\%$ <photon energy> errors requires
$$
\sigma_\alpha^2=2(0.01)^2,\qquad
\boxed{\sigma_\alpha=0.01414\ \mathrm{rad}\simeq0.81^\circ.}
$$
For independent equally precise single-<photon> polar directions, $\sigma_\alpha^2=\sigma_{\theta_1}^2+\sigma_{\theta_2}^2$ at this geometry, so
$$
\boxed{\sigma_\theta=0.0100\ \mathrm{rad}\simeq0.57^\circ\quad\text{per photon}.}
$$
An azimuthal error contributes only at second order about $\Delta\phi=\pi$. The corresponding <invariant-mass resolution> from the two <energy> errors alone is $0.707\%$; with an equally large angular contribution it is $1.00\%$. The common missing factor of four cancels when equating the two contributions, so the angular estimates remain valid despite the printed error.
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