Der Winkel$({75}^{\circ }+\delta )\backslash \; \backslash left(75^\backslash circ+\backslash delta\backslash right)\backslash$und der Winkel${130}^{\circ }\backslash \; 130^\backslash circ\backslash$sind Wechselwinkel,

Wechselwinkel sind gleich groß.

Der Winkel$\alpha \backslash \; \backslash alpha\backslash$und der Winkel$\delta \backslash \; \backslash delta\backslash$sind Scheitelwinkel,

Scheitelwinkel sind gleich groß.

Berechnung von$\delta \backslash \; \backslash delta$:

${75}^{\circ }+\delta ={130}^{\circ }\Rightarrow \delta ={130}^{\circ }-{75}^{\circ }={55}^{\circ }\backslash \; 75^\backslash circ+\backslash delta=130^\backslash circ\backslash Rightarrow\backslash delta=130^\backslash circ-75^\backslash circ=55^\backslash circ$

$\alpha =\delta \Rightarrow \boxed{{\displaystyle \mathbf{\alpha }={\mathbf{55}}^{\circ }}}\backslash \; \backslash alpha=\backslash delta\backslash Rightarrow\; \backslash boxed\{\backslash mathbf\{\backslash alpha=55^\backslash circ\}\}$

Der Winkel$\text{ EAB }\backslash \; \backslash text\{EAB\}\backslash$ist ein gestreckter Winkel$\Rightarrow \mathrm{\sphericalangle }\text{EAB}={180}^{\circ }\backslash \; \backslash Rightarrow\; \backslash sphericalangle\{\backslash text\{EAB\}\}=180^\backslash circ$

$\mathrm{\sphericalangle }\text{CAB}={180}^{\circ }-{130}^{\circ }={50}^{\circ }\backslash sphericalangle\{\backslash text\{CAB\}\}=180^\backslash circ-130^\backslash circ=50^\backslash circ$

Berechnung von $\mathbf{\beta }:\backslash mathbf\{\backslash beta\}:\backslash$

Das Dreieck$\text{ ABC }\backslash \; \backslash text\{ABC\}\backslash$ist ein gleichschenkliges Dreieck$\Rightarrow \mathrm{\sphericalangle }\mathrm{C}\mathrm{A}\mathrm{B}=\mathrm{\sphericalangle }\mathrm{A}\mathrm{B}\mathrm{C}\backslash \; \backslash Rightarrow\; \backslash mathrm\{\backslash sphericalangle\{CAB\}=\backslash sphericalangle\{ABC\}\}$

Die Winkelsumme im Dreieck ist ${180}^{\circ }\backslash \; 180^\backslash circ\backslash$

$\Rightarrow \mathrm{\sphericalangle }\mathrm{C}\mathrm{A}\mathrm{B}+\mathrm{\sphericalangle }\mathrm{A}\mathrm{B}\mathrm{C}+\beta ={180}^{\circ }\Rightarrow \beta ={180}^{\circ }-\mathrm{\sphericalangle }\mathrm{C}\mathrm{A}\mathrm{B}-\mathrm{\sphericalangle }\mathrm{A}\mathrm{B}\mathrm{C}\backslash Rightarrow\backslash \; \backslash mathrm\{\backslash sphericalangle\{CAB\}+\backslash sphericalangle\{ABC\}+\backslash beta=180^\backslash circ\}\backslash \; \backslash Rightarrow\; \backslash mathrm\{\backslash beta=180^\backslash circ-\backslash sphericalangle\{CAB\}-\backslash sphericalangle\{ABC\}\}$

$\beta ={180}^{\circ }-{50}^{\circ }-{50}^{\circ }\backslash hspace\{51mm\}\backslash mathrm\{\backslash beta=180^\backslash circ-50^\backslash circ-50^\backslash circ\}$

$\boxed{{\displaystyle \mathbf{\beta }={\mathbf{80}}^{\circ }}}\backslash hspace\{51mm\}\backslash boxed\{\backslash mathbf\{\backslash beta=80^\backslash circ\}\}$