%@Auteur: Christophe Poulain 
%@Date: 23 mai 2006 
\documentclass[12pt,a4paper]{article} 
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\usepackage{graphicx} 
\usepackage{calc} 
\usepackage{multicol} 
\usepackage{xlop} 
\usepackage{ifthen} 
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     \begin{lrbox}{\zouliboite} 
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     {\bf{\em Solution de l'exercice}}\par 
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\newboolean{exact} 
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\input{xlopsqrt} 
 
%% mise en scène d'un calcul 
\newcommand{\pythadroit}[5]{ 
    \opset{decimalsepsymbol={,}} 
    \opcopy{#4}{A1} 
    \opcopy{#5}{A2} 
    Dans le triangle $#1#2#3$ rectangle en $#2$, le théorème de 
    Pythagore permet d'écrire: 
    \begin{eqnarray*} 
 #1#3^2&=#1#2^2+#2#3^2\% 
 \opprint{A1}^2&=#1#2^2+\opprint{A2}^2\% 
 \opmul*{A1}{A1}{a1}\opprint{a1}&=#1#2^2+\opmul*{A2}{A2}{a2}\opprint{a2}\% 
 #1#2^2&=\opmul*{A1}{A1}{a1}\opprint{a1}-\opmul*{A2}{A2}{a2}\opprint{a2}\% 
 #1#2^2&=\opsub*{a1}{a2}{a3}\opprint{a3}\% 
 #1#2&=\sqrt{\opprint{a3}}\% 
 \ifthenelse{\boolean{exact}}% 
  {#1#2&=\opsqrt[maxdivstep=3]{a3}{a4}\opunzero{a4}\opprint{a4}}% 
  {#1#2&\approx\opsqrt[maxdivstep=3]{a3}{a4}\opround{a4}{2}{a4}\opunzero{a4}\opprint{a4}} 
    \end{eqnarray*} 
    } 
 
 
\pagestyle{empty} 
 
\begin{document} 
\textbf{Exercice} ---  Sur un cercle de centre $O$ et de diamètre $[AB]$ tel que $AB=10$~cm, 
place un point $C$ tel que l'angle $\widehat{ABC}=50\degres$. 
\begin{enumerate} 
\item Montre que le triangle $ABC$ est rectangle. 
\item Calcule les longueurs $BC$ et $AC$. (On donnera les valeurs 
arrondies au millimètre.) 
\end{enumerate} 
\begin{Solution} 
    \begin{enumerate} 
 \item Comme $C$ appartient au cercle de diamètre $[AB]$ alors le 
 triangle $ABC$ est rectangle en $C$. 
 \item 
     \begin{multicols}{2} 
      Dans le triangle $ABC$ rectangle en $C$, on a : 
  \begin{eqnarray*} 
      \cos\widehat{ABC}&=\frac{BC}{AB}\% 
      \cos50&=\frac{BC}{10}\% 
      BC&=10\times\cos50\% 
      BC&\approx6,4~\mbox{cm} 
  \end{eqnarray*} 
  \par\columnbreak\setboolean{exact}{false}\pythadroit ACB{10}{6,4} 
     \end{multicols} 
    \end{enumerate} 
\end{Solution} 
\end{document} 
 

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