Die Winkelsumme im Dreieck = ${180}^{\circ }180^\backslash circ$ also gilt:

$\alpha +\beta +\gamma ={180}^{\circ }\backslash alpha+\backslash beta+\backslash gamma\; =\; 180^\backslash circ$

Aus der Aufgabenstellung ist bekannt: $\alpha =3\gamma \beta =2\gamma \backslash hspace\{5mm\}\backslash alpha\; =\; 3\backslash gamma\; \backslash \backslash \backslash hspace\{57mm\}\backslash beta\; =\; 2\backslash gamma$

Jetzt kannst Du folgenden Ansatz bilden:

$3\gamma +2\gamma +\gamma ={180}^{\circ }6\gamma ={180}^{\circ }\mid :6\gamma ={30}^{\circ }3\backslash gamma+2\backslash gamma+\backslash gamma\; =\; 180^\backslash circ\backslash \backslash \backslash hspace\{14mm\}6\backslash gamma=\backslash hspace\{0.5mm\}180^\backslash circ\backslash hspace\{5mm\}\backslash big\backslash vert:6\backslash \backslash \backslash hspace\{16mm\}\backslash gamma=\backslash hspace\{1.5mm\}30^\backslash circ$

Der Winkel $\gamma \backslash gamma$ beträgt ${30}^{\circ }30^\backslash circ$.