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| author | Florian Fischer <florian.fischer@muhq.space> | 2024-06-07 15:28:30 +0200 |
|---|---|---|
| committer | Florian Fischer <florian.fischer@muhq.space> | 2024-06-07 15:28:30 +0200 |
| commit | a612d22755aef4e83e7d535d8df6121a8cfc96ec (patch) | |
| tree | 4200adcb6850eda193545d87eb8f8b6f9a0cef2d /Mathe/23_24/06 | |
| parent | c8e6c9f5bf581bd90acab6ed73f5e23c69b7e394 (diff) | |
| download | unterricht-a612d22755aef4e83e7d535d8df6121a8cfc96ec.tar.gz unterricht-a612d22755aef4e83e7d535d8df6121a8cfc96ec.zip | |
improve angle calculation exercises
Diffstat (limited to 'Mathe/23_24/06')
| -rw-r--r-- | Mathe/23_24/06/10/7A/test/Aufgaben/dreiecke_winkel_bestimmen.tex | 8 |
1 files changed, 4 insertions, 4 deletions
diff --git a/Mathe/23_24/06/10/7A/test/Aufgaben/dreiecke_winkel_bestimmen.tex b/Mathe/23_24/06/10/7A/test/Aufgaben/dreiecke_winkel_bestimmen.tex index a77a18e5..2552ae96 100644 --- a/Mathe/23_24/06/10/7A/test/Aufgaben/dreiecke_winkel_bestimmen.tex +++ b/Mathe/23_24/06/10/7A/test/Aufgaben/dreiecke_winkel_bestimmen.tex @@ -30,8 +30,8 @@ Hinweis: Der Punkt $M$ ist der Mittelpunkt des Kreises. %\draw pic["$\beta$", draw, angle radius=1cm] {angle=C--B--E}; %\draw pic["$\delta$", draw, angle radius=0.9cm] {angle=A--AD--B}; \draw pic[draw, angle radius=1.1cm] {angle=B--A--C}; -\draw pic["$\epsilon$", draw, angle radius=0.8cm] {angle=C--M--E}; -\draw pic["$\psi$", draw, angle radius=1.5cm] {angle=E--B--A}; +\draw pic["$\delta$", draw, angle radius=0.8cm] {angle=C--M--E}; +\draw pic["$\beta$", draw, angle radius=1.5cm] {angle=E--B--A}; \draw pic["$45\degree$", draw, angle radius=1.5cm] {angle=D--C--A}; \draw[Triangle-] ($(A) + (0.8cm, -0.1cm)$) -- ($(A) + (0.5cm, -1cm)$) node[below right, yshift=0.3cm] {$26\degree$}; \end{tikzpicture} @@ -43,9 +43,9 @@ Hinweis: Der Punkt $M$ ist der Mittelpunkt des Kreises. \begin{itemize} \item $\gamma = 90\degree$\Point (ACD Thaleskreis) \item $\alpha = 45\degree$\Point (Innenwinkelsumme ACD) -\item $\psi = 26\degree $\Point (AMB gleichschenklig) +\item $\beta = 26\degree $\Point (AMB gleichschenklig) %\item $\beta = 90\degree - 26\degree = 64\degree$\Point (ACB Thaleskreis) -\item $\epsilon = 180\degree - (180\degree - 2*26\degree) = 128\degree$\Point (Innenwinkelsumme gleichschenklig ABC, Scheitelwinkel) +\item $\delta = 180\degree - 2*26\degree = 128\degree$\Point (Innenwinkelsumme gleichschenklig ABC, Scheitelwinkel) % \item $\delta = 180\degree - 45\degree - (180\degree - 128\degree)) = 83\degree$\Point (Innenwinkelsumme AME, Scheitelwinkel) \end{itemize} \end{Lsg} |
