Lösung Winkel am Einheitskreis

Version 1.13 von Thomas Köhler am 2024/07/18 14:05

Winkel \(\alpha\)30°60°90°120°150°180°210°240°270°300°330°360°
\(\sin(\alpha)\)\(\frac{1}{2}\)\(\frac{\sqrt{3}{2}}\)\(1\)\(\frac{\sqrt{3}{2}\)\(\frac{1}{2}\)