If $\sqrt{2}\cos(A - B) = 1$ and $2\sin(A + B) = \sqrt{3}$ with $A, B$ acute, then $A = ?$

From $\sqrt2\cos(A-B)=1$, $\cos(A-B)=\tfrac{1}{\sqrt2}$, so $A-B=45^\circ$; from $2\sin(A+B)=\sqrt3$, $\sin(A+B)=\tfrac{\sqrt3}{2}$, so $A+B=60^\circ$. Adding the two equations gives $2A=105^\circ$, hence $A=52.5^\circ$.