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Given a cube $ABCDA'B'C'D'$ and S,T,R,N the middle of $AB, DD', DC,$ respectivley $BC'$. Find the cosine of the angle of $TS$ and $RN$. enter image description here

I tried to use paralelism and perpendicularity theorems, then the cosinus theorem, but I didn't match very well the reslts.

mmm
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  • How are you defining the angle between two non-intersecting line segments? – Parcly Taxel Nov 26 '20 at 18:49
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    Assuming that 'cosinus' means 'cosine', you can duplicate the cube and place it in front of the cube above. Then the new $R$ will coincide with $S$. The distances $TS, RN$ and the distance between $T$ and the new $N$ should be easy to find; finally apply cosine theorem. – player3236 Nov 26 '20 at 18:50
  • Oh, yes, that's an idea that leads very easy to the final result. – mmm Nov 26 '20 at 19:08

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