\ANG{
%
%%\item
%              
\begin{minipage}[t]{8 cm}
Die Zeichnung stellt die seitliche Wange einer Treppe dar, von der die
Ma"se $a=150 {\ \rm cm}$, $b=120 {\ \rm cm}$, $c=30 {\ \rm cm}$ und 
$d=30 {\ \rm cm}$ gegeben sind. Ihr Steigungswinkel $\varphi$ soll berechnet 
werden.
\end{minipage}
\begin{minipage}[t]{7 cm}
\beginpicture
\setcoordinatesystem units <0.25mm,0.25mm> point at 0 120
\accountingoff
\setplotarea x from 0 to 220, y from -15 to 80 

\plot  60  0   210  104.3   210  120    180 120     7.5 0   60 0 /
\plot  60  0   210    0     210  104.3 /
\setshadegrid span <0.9 mm>%
\vshade 7.5  0 0  <z,z,z,z> 60 0 36.5 <z,z,z,z> 180  83.5  120 <z,z,z,z> 210  104.3 120 /
\betweenarrows {$a$}  [0.375,0.75] from  60 -10    to 210   -10
\betweenarrows {$b$}  [0.375,0.75] from 220   0    to 220   120
%\betweenarrows {$c$}  [0.375,0.75] from 180 130     to 210   130
\arrow <1.8mm>  [0.375,0.75]         from 170  130  to 180   130
\arrow <1.8mm>  [0.375,0.75]         from 220  130  to 210   130
\put {$c$} at 195 130 
\arrow <1.8mm>  [0.375,0.75]         from 114.3 49.9 to 120    41.7
\arrow <1.8mm>  [0.375,0.75]         from 108.6 58.1 to 102.9  66.3
\put {$d$} at 111. 54.5 
\put {$A$} at  53  10
\put {$O$} at 202  10
\put {$B$} at 202 107.5
\put {$C$} at 182 110.5
\circulararc 34.8 degrees from 90 0 center at 60 0
\put {$\varphi$} at 82 7.5
\endpicture
\end{minipage}

\begin{enumerate}


\item
Berechne $|\overline{AC}|$ und $\angle OAC$.
\item
Berechne damit $\angle BAC$ und $\varphi$.
\end{enumerate}


\vskip 1 cm
}
\LSG{\begin{enumerate}
\item $|\overline{AC}|^2 = (a-c)^2 +b^2$ ergibt $|\overline{AC}| = 170 {\rm cm}$. 
$\tan{(\angle OAC)}=1$ also $\angle OAC= 45^o$. 
\item $\sin{(\angle BAC )}=\frac{\D d}{\D |\overline{AC}|}$ ergibt 
$\angle BAC =10{,}2 ^o$.
Daraus folgt $\varphi=34.8^o$.
\end{enumerate}
}
