 Solution

The angle $\theta$ between the oriented coordinate tangent vectors satisfies $\cos \theta =F/\sqrt{EG}$. The constant-angle assumption therefore gives $F=k\sqrt{E}\sqrt{G}$ for a constant $k$. The first fundamental form is positive definite, so $EG-F^2>0$ and hence $|k|<1$. Substitution into the geodesic coordinate grid identities from part (i) givesEliminating either derivative yields $(1-k^2)(\sqrt{E}{)}_v=0$ and $(1-k^2)(\sqrt{G}{)}_u=0$. Since $E,G>0$ and $1-k^2>0$, this provesIf the angle is understood between unoriented curves, its constancy initially fixes $|F|/\sqrt{EG}$; continuity fixes the sign on each connected chart when this quantity is nonzero, and the zero case already has $F=0$. Thus the same conclusion holds. In fact constant-angle geodesic coordinate grids are locally flat, although only the displayed derivatives are needed here.