Da die Geraden g und h parallel sind, sind $\beta \backslash beta$ und ${\beta }^{\mathrm{\prime }}\backslash beta\text{'}$ Wechselwinkel, Wechselwinkel sind gleich groß.

Es gilt: $\beta ={\beta }^{\mathrm{\prime }}={129}^{\circ }\backslash beta\backslash \; =\backslash \; \backslash beta\text{'}\backslash \; =\backslash \; 129^\{\backslash circ\}$

$\gamma und{\gamma }^{\mathrm{\prime }}\backslash gamma\backslash \; und\backslash \; \backslash gamma\text{'}\backslash$sind Scheitelwinkel, Scheitelwinkel sind gleich groß.

Es gilt: $\gamma ={\gamma }^{\mathrm{\prime }}\backslash gamma\backslash \; =\backslash \; \backslash gamma\text{'}$ = 90${}^{\circ }^\backslash circ$

$\beta und\delta \backslash beta\backslash \; und\backslash \; \backslash delta\backslash$sind Nebenwinkel, Nebenwinkel ergänzen sich zu 180${}^{\circ }^\backslash circ$.

Es gilt: $\beta +\delta \backslash beta\backslash \; +\backslash \; \backslash delta\backslash$= 180${}^{\circ }^\backslash circ$ $\to \backslash rightarrow$ $\delta =\backslash delta\backslash \; =$ 180${}^{\circ }^\backslash circ$ - 129${}^{\circ }^\backslash circ$ = 51${}^{\circ }^\backslash circ$

Aus der Skizze ist ersichtlich: $\alpha ={\gamma }^{\mathrm{\prime }}+\delta \backslash alpha\backslash \; =\backslash \; \backslash gamma\text{'}\backslash \; +\backslash \; \backslash delta$

Der gesuchte Winkel $\alpha ={90}^{\circ }+{51}^{\circ }={141}^{\circ }\backslash alpha\backslash \; =\backslash \; 90^\backslash circ\; +\; 51^\backslash circ\; =\; 141^\backslash circ$