 Solution

For two photons, four-momentum conservation and their vanishing rest masses give the diphoton invariant massTaking a logarithmic differential,For small mutually independent measurement errors, propagation of uncertainty consequently givesPrinted-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 $\mathrm{Var}(M^2)/M^4$ 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:At ${\theta }_1={\theta }_2=45^{\circ }$ and $\Delta \varphi =180^{\circ }$, $\cos \alpha =0$ and $\alpha =90^{\circ }$. The angular coefficient is therefore one. Equal contribution of opening-angle uncertainty and the combined two independent $1\%$ photon energy errors requiresFor independent equally precise single-photon polar directions, ${\sigma }_{\alpha }^2={\sigma }_{{\theta }_1}^2+{\sigma }_{{\theta }_2}^2$ at this geometry, soAn azimuthal error contributes only at second order about $\Delta \varphi =\pi$. 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.