We meet at infinity

One can observe side of nth square can be given by formula $\frac { \sin { \xi } }{ { \left( \sqrt { 2 } \right) }^{ n-1 } }$ which is very is easy to get.

Therefore area of nth square is given by ${ \left( \frac { \sin { \xi } }{ { \left( \sqrt { 2 } \right) }^{ n-1 } } \right) }^{ 2 }$ .

Observe that sequence of areas of squares forms a GP with first term $=a=\sin ^{ 2 }{ \xi }$ .

common ratio $=r=\frac { 1 }{ 2 }$ .

As $\left| r \right| <1$ we can apply infinite GP sum formula.

Therefore sum of areas of all square $=\frac { a }{ 1-r } =\frac { \sin ^{ 2 }{ \xi } }{ 1-\frac { 1 }{ 2 } } =2\sin ^{ 2 }{ \xi }$

It is given that $2\sin ^{ 2 }{ \xi } =\frac { 3 }{ 2 }$

$\Rightarrow \sin { \xi } =\frac { \sqrt { 3 } }{ 2 } \Rightarrow \xi =60°$ .