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

$\begin{array}{l}\alpha +\beta +{100}^{\circ }={180}^{\circ }\\ \alpha +\beta ={80}^{\circ }\end{array}$

Die Summe von Nebenwinkeln beträgt ${180}^{\circ }$.

$\begin{array}{l}\alpha +\alpha \ast ={180}^{\circ }\\ \alpha ={180}^{\circ }-\alpha \ast \\ \\ \beta +\beta \ast ={180}^{\circ }\\ \beta ={180}^{\circ }-\beta \ast \end{array}$

$\begin{array}{l}({180}^{\circ }-\alpha \ast )+({180}^{\circ }-\beta \ast )={80}^{\circ }\end{array}$

$\begin{array}{l}{360}^{\circ }-\alpha \ast -\beta \ast ={80}^{\circ }\\ \end{array}$

${280}^{\circ }=\alpha \ast +\beta \ast$

Die Lösung lautet $\alpha \ast +\beta \ast ={280}^{\circ }$