\documentclass[a4paper,11pt]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[french]{babel}
\usepackage[a4paper,top=3.2cm,bottom=2cm,left=2cm,right=2cm]{geometry}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\usepackage{fancyhdr}
\usepackage{qrcode}
\usepackage[table]{xcolor}
\usepackage{amsmath,amssymb}
\usepackage{enumitem}
% Puces nettes (babel-french remplace sinon la puce par un tiret « – » peu pro).
% Vaut pour TOUTES les fiches qui importent ce module. AtBeginDocument pour
% passer après la personnalisation des listes faite par babel-french.
\AtBeginDocument{%
  \setlist[itemize]{leftmargin=1.8em,itemsep=0.5ex,topsep=0.4ex,parsep=0pt}%
  \setlist[itemize,1]{label=\textcolor{red!75!black}{\small\textbullet}}%
  \setlist[itemize,2]{label=\textcolor{red!75!black}{\scriptsize\textbullet}}%
}
% Titres compacts (gain de place vertical, fiches plus denses) — vaut pour tous.
\usepackage{titlesec}
\titlespacing*{\section}{0pt}{1.4ex}{0.6ex}
\titlespacing*{\subsection}{0pt}{1.0ex}{0.3ex}
% multicol : listes denses (calcul mental) sur 2 colonnes — neutre tant qu'il
% n'est pas utilisé, dispo pour tous les générateurs.
\usepackage{multicol}
\setlength{\columnsep}{1.5em}
% needspace : empêche un titre d'exercice de rester orphelin en bas de page
% (réserve quelques lignes ; sinon le titre passe à la page suivante avec sa
% liste). Neutre tant qu'il n'est pas appelé — dispo pour tous les générateurs.
\usepackage{needspace}
\usepackage{tikz}
\usetikzlibrary{decorations.text}

\usepackage{tgheros}
\pagestyle{fancy}
\fancyhf{}
\setlength{\headheight}{75pt}
\setlength{\headsep}{10pt}

\newcommand{\logowouf}{%
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=0.55,
                    line cap=round,line join=round]
    \path[use as bounding box] (-2.20,-1.45) rectangle (2.20,2.20);
    \path[
        postaction={decorate},
        decoration={
            text along path,
            text={|\fontfamily{qhv}\bfseries\fontsize{9}{9}\selectfont|site2wouf.fr},
            text align={center},
            raise=0.03ex
        }
    ] (-1.29,0.48) arc[start angle=160,end angle=20,radius=1.29];
    \begin{scope}[xshift=-0.06cm]
        \fill[black] (-0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] ( 0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] (-0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black] ( 0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black]
            (0,-0.10)
            .. controls (-0.36,-0.02) and (-0.63,-0.24) .. (-0.58,-0.58)
            .. controls (-0.52,-0.93) and (-0.16,-1.12) .. (0,-0.90)
            .. controls (0.16,-1.12) and (0.52,-0.93) .. (0.58,-0.58)
            .. controls (0.63,-0.24) and (0.36,-0.02) .. (0,-0.10)
            -- cycle;
    \end{scope}
\end{tikzpicture}%
}

\newcommand{\titre}{Trigonométrie}
\newcommand{\soustitre}{Cycle 4 --- Fiche n\textdegree 135}

\fancyhead[C]{%
\raisebox{4pt}{%
\begin{minipage}[c]{0.15\textwidth}
    \centering
    \logowouf
\end{minipage}
\begin{minipage}[c]{0.62\textwidth}
    \centering
    {\Large\bfseries \titre\par}
    {\small\color{gray} \soustitre}
\end{minipage}
\begin{minipage}[c]{0.21\textwidth}
    \centering
    \qrcode[height=1.5cm]{https://site2wouf.fr/exercices_trigonometrie.php?x=135}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0135}
\end{minipage}%
}
}

\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle VDP rectangle en V, on sait que :
\begin{itemize}
\item VP=4,7 cm
\item DP=8,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{VPD}\).

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle GZH rectangle en G, on sait que :
\begin{itemize}
\item GH=4,4 cm
\item \(\widehat{ZHG}\) =  43\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [HZ]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle GBJ rectangle en G, on sait que :
\begin{itemize}
\item BJ=2,1 cm
\item \(\widehat{GBJ}\) = 59\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [GJ]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle BTS rectangle en B, on sait que :
\begin{itemize}
\item BT=2,4 cm
\item BS=5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BTS}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle CWB rectangle en C, on sait que :
\begin{itemize}
\item CB=8,7 cm
\item \(\widehat{WBC}\) =  31\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [CW]. (Arrondir au dixième)

\newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.571,3.400) -- (2.578,3.050) -- (2.558,2.701) -- (2.565,2.351) -- (2.575,2.002) -- (2.562,1.647) -- (2.578,1.299) -- (2.565,0.949) -- (2.571,0.600);
  \draw[black] (2.571,0.600) -- (2.801,0.603) -- (3.035,0.594) -- (3.270,0.602) -- (3.498,0.609) -- (3.736,0.605) -- (3.965,0.600) -- (4.196,0.592) -- (4.429,0.600);
  \draw[black] (4.429,0.600) -- (4.235,0.873) -- (4.047,1.155) -- (3.879,1.444) -- (3.677,1.711) -- (3.493,1.993) -- (3.316,2.278) -- (3.128,2.559) -- (2.936,2.833) -- (2.760,3.123) -- (2.571,3.400);
  \draw[black] (2.571,0.786) -- (2.575,0.724) -- (2.566,0.662) -- (2.571,0.600);
  \draw[black] (2.571,0.600) -- (2.634,0.596) -- (2.694,0.601) -- (2.757,0.600);
  \draw[black] (2.571,0.786) -- (2.633,0.792) -- (2.695,0.780) -- (2.757,0.786);
  \draw[black] (2.757,0.786) -- (2.759,0.725) -- (2.753,0.663) -- (2.757,0.600);
  \draw[black] (4.206,0.935) -- (4.178,0.918) -- (4.155,0.894) -- (4.128,0.874) -- (4.119,0.839) -- (4.100,0.813) -- (4.068,0.794) -- (4.062,0.760) -- (4.055,0.728) -- (4.036,0.699) -- (4.027,0.667) -- (4.034,0.633) -- (4.035,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.304,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.304,3.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.529,0.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.842,0.849) {?};
\end{tikzpicture}\end{center}


 Dans le triangle VDP rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VPD}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{VPD}\)) = \(\dfrac{VP}{DP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{VPD}\)) =  \(\dfrac{4{,}7}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{VPD}\) = Arccos (\(\dfrac{4{,}7}{8{,}5}\)) \(\approx\) 56\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.999,3.400) -- (1.995,3.051) -- (1.991,2.702) -- (1.987,2.351) -- (1.991,1.997) -- (2.010,1.652) -- (1.999,1.299) -- (2.007,0.948) -- (1.999,0.600);
  \draw[black] (1.999,0.600) -- (2.332,0.605) -- (2.664,0.598) -- (2.997,0.607) -- (3.333,0.608) -- (3.669,0.596) -- (3.998,0.602) -- (4.332,0.600) -- (4.668,0.601) -- (5.001,0.600);
  \draw[black] (5.001,0.600) -- (4.745,0.830) -- (4.505,1.068) -- (4.244,1.289) -- (4.010,1.541) -- (3.742,1.757) -- (3.499,2.001) -- (3.245,2.231) -- (3.008,2.475) -- (2.751,2.697) -- (2.493,2.925) -- (2.255,3.171) -- (1.999,3.400);
  \draw[black] (1.999,0.880) -- (1.998,0.788) -- (1.995,0.693) -- (1.999,0.600);
  \draw[black] (1.999,0.600) -- (2.092,0.600) -- (2.185,0.602) -- (2.279,0.600);
  \draw[black] (1.999,0.880) -- (2.091,0.874) -- (2.185,0.883) -- (2.279,0.880);
  \draw[black] (2.279,0.880) -- (2.276,0.788) -- (2.278,0.692) -- (2.279,0.600);
  \draw[black] (4.704,0.878) -- (4.693,0.854) -- (4.679,0.832) -- (4.661,0.815) -- (4.657,0.788) -- (4.631,0.773) -- (4.635,0.744) -- (4.618,0.724) -- (4.608,0.700) -- (4.614,0.674) -- (4.607,0.650) -- (4.596,0.625) -- (4.599,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.732,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.732,3.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.101,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.381,0.778) {43°};
\end{tikzpicture}\end{center}


 Dans le triangle GZH rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GHZ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{GHZ}\)) = \(\dfrac{GH}{ZH}\)
\par\addvspace{.5em}
  d'où    cos(43\textdegree{}) =  \(\dfrac{4{,}4}{ZH}\)
\par\addvspace{.5em}
 On a donc ZH =  \(\dfrac{4{,}4}{\cos(43^{\circ})}\) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.170,3.400) -- (1.169,3.053) -- (1.177,2.702) -- (1.161,2.349) -- (1.180,1.997) -- (1.157,1.652) -- (1.180,1.303) -- (1.182,0.950) -- (1.170,0.600);
  \draw[black] (1.170,0.600) -- (1.528,0.601) -- (1.884,0.600) -- (2.247,0.590) -- (2.601,0.608) -- (2.966,0.600) -- (3.321,0.609) -- (3.680,0.589) -- (4.037,0.604) -- (4.396,0.610) -- (4.753,0.611) -- (5.116,0.600) -- (5.469,0.587) -- (5.830,0.600);
  \draw[black] (5.830,0.600) -- (5.533,0.765) -- (5.247,0.950) -- (4.957,1.131) -- (4.660,1.291) -- (4.365,1.466) -- (4.088,1.654) -- (3.797,1.829) -- (3.504,2.010) -- (3.209,2.176) -- (2.915,2.344) -- (2.625,2.527) -- (2.344,2.710) -- (2.045,2.871) -- (1.745,3.042) -- (1.462,3.228) -- (1.170,3.400);
  \draw[black] (1.170,0.880) -- (1.175,0.785) -- (1.168,0.694) -- (1.170,0.600);
  \draw[black] (1.170,0.600) -- (1.263,0.603) -- (1.358,0.605) -- (1.450,0.600);
  \draw[black] (1.170,0.880) -- (1.262,0.878) -- (1.355,0.882) -- (1.450,0.880);
  \draw[black] (1.450,0.880) -- (1.456,0.786) -- (1.444,0.695) -- (1.450,0.600);
  \draw[black] (1.521,3.189) -- (1.499,3.162) -- (1.468,3.142) -- (1.445,3.118) -- (1.423,3.092) -- (1.398,3.067) -- (1.368,3.051) -- (1.333,3.045) -- (1.302,3.030) -- (1.271,3.018) -- (1.239,3.002) -- (1.204,3.002) -- (1.170,3.005);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.903,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.903,3.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.930,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.498,2.753) {59°};
\end{tikzpicture}\end{center}


 Dans le triangle GBJ rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GBJ}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{GBJ}\)) =  \(\dfrac{GJ}{BJ}\)
\par\addvspace{.5em}
 d'où sin(59\textdegree{}) = \(\dfrac{GJ}{2{,}1}\)

On a donc GJ = 2,1 \(\times\) sin(59\textdegree{}) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.176) -- (1.042,2.882) -- (1.041,2.588) -- (1.054,2.292) -- (1.044,2.001) -- (1.052,1.709) -- (1.061,1.409) -- (1.060,1.119) -- (1.050,0.824);
  \draw[black] (1.050,0.824) -- (1.398,0.828) -- (1.751,0.837) -- (2.100,0.830) -- (2.450,0.821) -- (2.800,0.825) -- (3.150,0.818) -- (3.497,0.829) -- (3.847,0.835) -- (4.199,0.837) -- (4.548,0.835) -- (4.898,0.817) -- (5.250,0.837) -- (5.599,0.812) -- (5.950,0.824);
  \draw[black] (5.950,0.824) -- (5.641,0.961) -- (5.340,1.126) -- (5.030,1.271) -- (4.723,1.407) -- (4.422,1.568) -- (4.117,1.718) -- (3.807,1.860) -- (3.497,1.989) -- (3.197,2.148) -- (2.889,2.294) -- (2.588,2.449) -- (2.271,2.586) -- (1.970,2.744) -- (1.662,2.885) -- (1.351,3.017) -- (1.050,3.176);
  \draw[black] (1.050,1.059) -- (1.052,0.980) -- (1.052,0.902) -- (1.050,0.824);
  \draw[black] (1.050,0.824) -- (1.127,0.822) -- (1.207,0.820) -- (1.285,0.824);
  \draw[black] (1.050,1.059) -- (1.130,1.054) -- (1.207,1.053) -- (1.285,1.059);
  \draw[black] (1.285,1.059) -- (1.288,0.981) -- (1.283,0.903) -- (1.285,0.824);
  \draw[black] (1.419,2.999) -- (1.390,2.971) -- (1.368,2.942) -- (1.350,2.908) -- (1.322,2.884) -- (1.296,2.855) -- (1.258,2.845) -- (1.232,2.817) -- (1.198,2.799) -- (1.159,2.798) -- (1.124,2.784) -- (1.088,2.769) -- (1.050,2.781);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.691) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.243) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.691) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.491) {5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.405,2.545) {?};
\end{tikzpicture}\end{center}


 Dans le triangle BTS rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BTS}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{BTS}\)) = \(\dfrac{BS}{BT}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BTS}\)) = \(\dfrac{5}{2{,}4}\)
\par\addvspace{.5em}
 On a donc BTS = ArcTan(\(\dfrac{5}{2{,}4}\)) \(\approx\) 64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.170,3.400) -- (1.169,3.051) -- (1.161,2.699) -- (1.167,2.348) -- (1.181,1.999) -- (1.179,1.651) -- (1.177,1.301) -- (1.172,0.950) -- (1.170,0.600);
  \draw[black] (1.170,0.600) -- (1.529,0.605) -- (1.890,0.604) -- (2.243,0.603) -- (2.602,0.596) -- (2.963,0.606) -- (3.321,0.588) -- (3.678,0.609) -- (4.039,0.601) -- (4.398,0.594) -- (4.754,0.604) -- (5.110,0.607) -- (5.474,0.592) -- (5.830,0.600);
  \draw[black] (5.830,0.600) -- (5.544,0.781) -- (5.243,0.948) -- (4.956,1.127) -- (4.662,1.292) -- (4.377,1.487) -- (4.080,1.647) -- (3.795,1.833) -- (3.506,2.010) -- (3.213,2.179) -- (2.912,2.344) -- (2.622,2.512) -- (2.344,2.709) -- (2.044,2.881) -- (1.756,3.051) -- (1.467,3.236) -- (1.170,3.400);
  \draw[black] (1.170,0.880) -- (1.177,0.788) -- (1.175,0.693) -- (1.170,0.600);
  \draw[black] (1.170,0.600) -- (1.264,0.599) -- (1.357,0.599) -- (1.450,0.600);
  \draw[black] (1.170,0.880) -- (1.263,0.880) -- (1.358,0.876) -- (1.450,0.880);
  \draw[black] (1.450,0.880) -- (1.452,0.787) -- (1.450,0.694) -- (1.450,0.600);
  \draw[black] (5.495,0.801) -- (5.475,0.792) -- (5.465,0.776) -- (5.459,0.759) -- (5.457,0.741) -- (5.459,0.721) -- (5.449,0.706) -- (5.439,0.690) -- (5.434,0.672) -- (5.440,0.653) -- (5.437,0.636) -- (5.435,0.618) -- (5.430,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.903,0.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.903,3.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.930,0.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.670,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {8,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.188,0.711) {31°};
\end{tikzpicture}\end{center}


 Dans le triangle CWB rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CBW}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{CBW}\)) = \(\dfrac{CW}{CB}\)
\par\addvspace{.5em}
d'où  tan(31\textdegree{}) =  \(\dfrac{CW}{8{,}7}\)
\par\addvspace{.5em}
On a donc CW =  8,7 \(\times\) tan(31\textdegree{})  \(\approx\) 5,2 cm\end{document}
