\documentclass[a4paper,11pt]{article} \usepackage{francois_meria} \usepackage[dvips]{graphicx} \usepackage[dvips]{epsfig} \usepackage{calc} \setlength{\parindent}{0mm} \lhead{\textsf{Collège Château Forbin} - \textit{Mathématiques} - \textsf{4\ieme 1}} \chead{} \rhead{\textit{Année} 2005/2006} \pagestyle{fancy} \renewcommand{\headrulewidth}{0.5pt} \begin{document} \centerline{\LARGE Triangle RECTANGLE : une nouvelle notion} \vskip 2cm On considère un triangle rectangle. Le but de ces exercices est de découvrir une nouvelle propriété du triangle rectangle qui sert à calculer des \og \textit{longueurs} \fg~ manquantes dans un tel triangle. \vskip 2cm \begin{multicols}{2} \begin{center} \psset{unit=1.5cm} \pspicture(4,2.25) \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){A}(4,0){B} \pstGeonode[PointSymbol=none,PointName=none](0,1){T} \pstRotation[PointSymbol=none,RotAngle=-30,PointName=none]{B}{A}{R} \pstInterLL[PointSymbol=none,PosAngle=90]{A}{T}{B}{R}{C} \pstLineAB{A}{C} \pstLineAB{A}{B} \pstLineAB{B}{C} \pstRightAngle[RightAngleSize=0.2]{B}{A}{C} \pstMarkAngle{C}{B}{A}{30°} \pcline{<->}(0,-0.25)(4,-0.25) \lput*{:U}{4~cm} \put(3,2){1\ier ~triangle} \endpspicture \end{center} \begin{center} \psset{unit=1.5cm} \pspicture(4,2.25) \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){A}(4,0){B} \pstGeonode[PointSymbol=none,PointName=none](0,1){T} \pstRotation[PointSymbol=none,RotAngle=-30,PointName=none]{B}{A}{R} \pstInterLL[PointSymbol=none,PosAngle=90]{A}{T}{B}{R}{C} \pstLineAB{A}{C} \pstLineAB{A}{B} \pstLineAB{B}{C} \pstRightAngle[RightAngleSize=0.2]{B}{A}{C} \pstMarkAngle{C}{B}{A}{30°} \pcline{<->}(0,-0.25)(4,-0.25) \lput*{:U}{6~cm} \put(3,2){2\ieme ~triangle} \endpspicture \end{center} \end{multicols} \vskip 1.5cm Reproduire en vraies grandeurs les deux triangles $ABC$ ci-dessus, puis compléter le tableau suivant. \vskip 1.5cm \begin{center} \begin{tabularx}{\textwidth}{|X|X|X|X|} \cline{2-4} \multicolumn{1}{c|}{} & & & \\ \multicolumn{1}{c|}{} & Mesure de $BA$ & Mesure de $BC$ & Rapport $\dfrac{BA}{BC}$ \\ \multicolumn{1}{c|}{} & & & \\ \hline & & & \\ {\Large 1\ier ~ triangle} & & & \\ & & & \\ \hline & & & \\ {\Large2\ieme ~ triangle} & & & \\ & & & \\ \hline \end{tabularx} \end{center} \vskip 0.5cm Que remarque-t-on ? \dotfill\\ \null \dotfill\\ \null \dotfill\\ \vskip 0.2cm \textbf{Définition : } \dotfill\\ \null \dotfill\\ \null \dotfill\\ \null \dotfill\\\null \dotfill\\ \newpage \begin{exercice} En utilisant les figures suivantes, compléter le tableau ci-dessous. \vskip 1.5cm \begin{multicols}{3} \begin{tabular}{ll} \pspicture(0,-1)(2.5,1) \put(0,0.5){\begin{tabular}{l}$AB=4$~cm \\ $BC=6$~cm \end{tabular}} \endpspicture & \psset{unit=1cm} \pspicture(0,-1)(2.5,1) \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){A}(2.5,0){B} \pstGeonode[PointSymbol=none,PointName=none](0,0.5){T} \pstRotation[RotAngle=-25,PointName=none,PointSymbol=none]{B}{A}{S} \pstInterLL[PointSymbol=none,PosAngle=90]{A}{T}{B}{S}{C} \pstRightAngle[RightAngleSize=0.2]{B}{A}{C} \pstLineAB{A}{B} \pstLineAB{A}{C} \pstLineAB{B}{C} \pstMarkAngle[MarkAngleRadius=0.8]{C}{B}{A}{} \endpspicture \\ \end{tabular} \begin{tabular}{ll} \pspicture(0,-1)(2.5,1) \put(0,0.5){\begin{tabular}{l}$WN=6$~cm \\ $NA=3$~cm \end{tabular}} \endpspicture & \psset{unit=1cm} \pspicture(0,-1)(2.5,1) \rput{-5}{ \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){A}(2.5,0){W} \pstGeonode[PointSymbol=none,PointName=none](0,0.5){T} \pstRotation[RotAngle=-25,PointName=none,PointSymbol=none]{W}{A}{S} \pstInterLL[PointSymbol=none,PosAngle=90]{A}{T}{W}{S}{N} \pstRightAngle[RightAngleSize=0.2]{W}{A}{N} \pstLineAB{A}{W} \pstLineAB{A}{N} \pstLineAB{W}{N} \pstMarkAngle[MarkAngleRadius=0.5]{A}{N}{W}{} } \endpspicture \\ \end{tabular} \begin{tabular}{ll} \pspicture(0,-1)(2.5,1) \put(0,0.5){\begin{tabular}{l}$FK=3$~cm \\ $FD=4$~cm \end{tabular}} \endpspicture & \psset{unit=1cm} \pspicture(0,-1)(2.5,1) \rput{5}{ \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){F}(2.5,0){D} \pstGeonode[PointSymbol=none,PointName=none](0,0.5){T} \pstRotation[RotAngle=-25,PointName=none,PointSymbol=none]{D}{F}{S} \pstInterLL[PointSymbol=none,PosAngle=90]{F}{T}{D}{S}{K} \pstRightAngle[RightAngleSize=0.2]{D}{F}{K} \pstLineAB{F}{D} \pstLineAB{F}{K} \pstLineAB{D}{K} \pstMarkAngle[MarkAngleRadius=0.8]{K}{D}{F}{} } \endpspicture \\ \end{tabular} \end{multicols} Penser au théorème de Pythagore dans le troisième exemple. \vskip 0.5cm \begin{center} \begin{tabularx}{\textwidth}{|>{\centering}X|>{\centering}X|>{\centering}X|>{\centering}X|} \cline{2-4} \multicolumn{1}{c|}{} & & & \\ \multicolumn{1}{c|}{} & Triangle $ABC$ & Triangle $AWN$& Triangle $FDK$ \tabularnewline \multicolumn{1}{c|}{} & & & \tabularnewline \hline longueur du & & & \tabularnewline côté adjacent de & 4~cm & & \tabularnewline l'angle marqué & & & \tabularnewline \hline longueur de & & & \tabularnewline l'hypoténuse du & 6~cm & & \tabularnewline triangle & & & \tabularnewline \hline & & & \tabularnewline $\dfrac{\textrm{côté adjacent}}{\textrm{hypoténuse}}$ & $\dfrac{4~\textrm{cm}}{6~\textrm{cm}}=\dfrac{2}{3}\approx 0,67$ & & \tabularnewline & & & \tabularnewline \hline cosinus de & & & \tabularnewline l'angle marqué & $0,67$ & & \tabularnewline $\approx$ & & & \tabularnewline \hline valeur (en °) & & & \tabularnewline de l'angle & $48,2$° & & \tabularnewline marqué & & & \tabularnewline \hline \end{tabularx} \end{center} À l'aide de la calculatrice, remplir les tableaux suivants :\\ % Table generated by Excel2LaTeX from sheet 'Feuil1' \begin{tabularx}{\textwidth}{|c*{11}{|>{\centering}X}|} \hline {\bf angle en °} & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 \tabularnewline \hline {\bf } & & & & & & & & & & & \tabularnewline {\bf cosinus} & & & & & & & & & & & \tabularnewline & & & & & & & & & & & \tabularnewline \hline \end{tabularx} \vskip 0.5cm % Table generated by Excel2LaTeX from sheet 'Feuil1' \begin{tabularx}{\textwidth}{|c*{8}{|>{\centering}X}|} \hline {\bf angle en °} & 55 & 60 & 65 & 70 & 75 & 80 & 85 & 89 \tabularnewline \hline {\bf } & & & & & & & & \tabularnewline {\bf cosinus} & & & & & & & & \tabularnewline & & & & & & & & \tabularnewline \hline \end{tabularx} \vskip 0.5cm % Table generated by Excel2LaTeX from sheet 'Feuil1' \begin{tabularx}{\textwidth}{|c*{9}{|>{\centering}X}|} \hline {\bf } & & & & & & & & & \tabularnewline {\bf angle en °} & & & & & & & & & \tabularnewline & & & & & & & & & \tabularnewline \hline {\bf cosinus} & 0,9 & 0,8 & 0,7 & 0,6 & 0,5 & 0,4 & 0,3 & 0,2 & 0,1 \tabularnewline \hline \end{tabularx} \end{exercice} En utilisant les données de chaque figure, calculer la valeur manquante : \vskip 1.5cm \begin{multicols}{3} \begin{tabular}{ll} \pspicture(0,-1)(2.5,1) \put(0,0.5){\begin{tabular}{l}$AB=$ ?~cm \\ $BC=6$~cm \end{tabular}} \endpspicture & \psset{unit=1cm} \pspicture(0,-1)(2.5,1) \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){A}(2.5,0){B} \pstGeonode[PointSymbol=none,PointName=none](0,0.5){T} \pstRotation[RotAngle=-25,PointName=none,PointSymbol=none]{B}{A}{S} \pstInterLL[PointSymbol=none,PosAngle=90]{A}{T}{B}{S}{C} \pstRightAngle[RightAngleSize=0.2]{B}{A}{C} \pstLineAB{A}{B} \pstLineAB{A}{C} \pstLineAB{B}{C} \pstMarkAngle[MarkAngleRadius=0.8]{C}{B}{A}{50°~~} \endpspicture \\ \end{tabular} \begin{tabular}{ll} \pspicture(0,-1)(2.5,1) \put(0,0.5){\begin{tabular}{l}$WN=$~?~cm \\ $NA=3$~cm \end{tabular}} \endpspicture & \psset{unit=1cm} \pspicture(0,-1)(2.5,1) \rput{-5}{ \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){A}(2.5,0){W} \pstGeonode[PointSymbol=none,PointName=none](0,0.5){T} \pstRotation[RotAngle=-25,PointName=none,PointSymbol=none]{W}{A}{S} \pstInterLL[PointSymbol=none,PosAngle=90]{A}{T}{W}{S}{N} \pstRightAngle[RightAngleSize=0.2]{W}{A}{N} \pstLineAB{A}{W} \pstLineAB{A}{N} \pstLineAB{W}{N} \pstMarkAngle[MarkAngleRadius=0.4]{A}{N}{W}{40°} } \endpspicture \\ \end{tabular} \begin{tabular}{ll} \pspicture(0,-1)(2.5,1) \put(0,0.5){\begin{tabular}{l}$KD=10$~cm \\ $FD=7$~cm \end{tabular}} \endpspicture & \psset{unit=1cm} \pspicture(0,-1)(2.5,1) \rput{5}{ \pstGeonode[PointSymbol=none,PosAngle={235,-45}](0,0){F}(2.5,0){D} \pstGeonode[PointSymbol=none,PointName=none](0,0.5){T} \pstRotation[RotAngle=-25,PointName=none,PointSymbol=none]{D}{F}{S} \pstInterLL[PointSymbol=none,PosAngle=90]{F}{T}{D}{S}{K} \pstRightAngle[RightAngleSize=0.2]{D}{F}{K} \pstLineAB{F}{D} \pstLineAB{F}{K} \pstLineAB{D}{K} \pstMarkAngle[MarkAngleRadius=0.8]{K}{D}{F}{~?~~} } \endpspicture \\ \end{tabular} \end{multicols} \end{document}