Bestimmung des Winkels $\alpha \backslash large\{\backslash alpha\}$:

Der Winkel ${140}^{\circ }140^\backslash circ$ und der Winkel $ϵ\backslash large\{\backslash epsilon\}$ bilden einen gestreckten Winkel, ein gestreckter Winkel hat eine Größe von ${180}^{\circ }180^\backslash circ$.

Es gilt: ${140}^{\circ }+ϵ={180}^{\circ }\Rightarrow ϵ={180}^{\circ }-{140}^{\circ }ϵ={40}^{\circ }140^\backslash circ\backslash \; +\backslash \; \{\backslash large\{\backslash epsilon\}\}\backslash \; =\backslash \; 180^\backslash circ\backslash qquad\backslash Rightarrow\backslash qquad\{\backslash large\{\backslash epsilon\}\}\backslash \; =\backslash \; 180^\backslash circ\backslash \; -\backslash \; 140^\backslash circ\backslash \backslash \backslash hspace\{23\; mm\}\{\backslash large\{\backslash epsilon\}\}\backslash \; =\backslash \; 40^\backslash circ$

Die Winkelsumme im Dreieck beträgt ${180}^{\circ }180^\backslash circ$.

Also ist: ${40}^{\circ }+{120}^{\circ }+\alpha ={180}^{\circ }\Rightarrow \alpha ={180}^{\circ }-{160}^{\circ }\alpha ={20}^{\circ }\backslash \; 40^\backslash circ\backslash \; +\backslash \; 120^\backslash circ\backslash \; +\backslash \; \{\backslash large\backslash alpha\}\; =\backslash \; 180^\backslash circ\backslash qquad\backslash Rightarrow\backslash qquad\backslash \; \{\backslash large\backslash alpha\}\backslash \; =\backslash \; 180^\backslash circ\backslash \; -\backslash \; 160^\backslash circ\backslash \backslash \backslash hspace\{36\; mm\}\backslash large\backslash alpha\backslash \; =\backslash \; 20^\backslash circ$

Bestimmung des Winkels $\beta \backslash large\{\backslash beta\}$:

Bestimme zuerst die Winkel des Dreiecks $\text{ACE}\backslash text\{ACE\}$

Die Winkel ${120}^{\circ }120^\backslash circ$ und der Winkel $\gamma \backslash large\{\backslash gamma\}$ sind Scheitelwinkel; Scheitelwinkel sind gleich groß,

also ist:

$\gamma ={120}^{\circ }\backslash gamma\backslash \; =\backslash \; 120^\backslash circ$

Das Dreieck $\text{ACE}\backslash text\{ACE\}$ ist ein gleichschenkliges Dreieck $\Rightarrow {\delta }_A={\delta }_B\backslash quad\backslash Rightarrow\backslash quad\; \backslash delta\_\{\backslash small\{A\}\}\backslash \; =\backslash \; \backslash delta\_\{\backslash small\{B\}\}$

${\delta }_A+{\delta }_B={180}^{\circ }-{120}^{\circ }={60}^{\circ }{\delta }_A={\delta }_B={30}^{\circ }\backslash delta\_\{\backslash small\{A\}\}\backslash \; +\backslash \; \backslash delta\_\{\backslash small\{B\}\}\backslash \; =\backslash \; 180^\backslash circ\backslash \; -\backslash \; 120^\backslash circ\backslash \; =\backslash \; 60^\backslash circ\backslash hspace\{35mm\}\backslash delta\_\{\backslash small\{A\}\}\backslash \; =\backslash \; \backslash delta\_\{\backslash small\{B\}\}\backslash \; =\backslash \; 30^\backslash circ$

Jetzt kannst Du den Winkel $\beta \backslash large\{\backslash beta\}\backslash$bestimmen.

Der Winkel $\beta \backslash large\{\backslash beta\}\backslash$und der Winkel ${\delta }_A\{\backslash large\{\backslash delta\}\}\_\{\backslash small\{A\}\}\backslash$sind Wechselwinkel, Wechselwinkel (auch Z-Winkel genannt) an parallelen Geraden sind gleich groß.

$\beta ={30}^{\circ }\backslash hspace\{36\; mm\}\backslash large\{\backslash beta\}\backslash \; =\backslash \; 30^\backslash circ$