\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 134}

\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=134}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0134}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle LCS rectangle en L, on sait que :
\begin{itemize}
\item LS=9,9 cm
\item \(\widehat{LCS}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle HJD rectangle en H, on sait que :
\begin{itemize}
\item HJ=1,5 cm
\item HD=4,2 cm
\end{itemize}

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

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

 Dans le triangle BTH rectangle en B, on sait que :
\begin{itemize}
\item TH=1,6 cm
\item \(\widehat{BTH}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle CWN rectangle en C, on sait que :
\begin{itemize}
\item CW=2,7 cm
\item WN=9,3 cm
\end{itemize}

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

 Dans le triangle NTJ rectangle en N, on sait que :
\begin{itemize}
\item NJ=2,6 cm
\item \(\widehat{TJN}\) =  15\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.254,3.051) -- (1.248,2.698) -- (1.266,2.351) -- (1.247,1.997) -- (1.249,1.651) -- (1.254,1.299) -- (1.265,0.951) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.607,0.605) -- (1.947,0.612) -- (2.295,0.595) -- (2.636,0.588) -- (2.981,0.589) -- (3.327,0.606) -- (3.674,0.608) -- (4.019,0.609) -- (4.364,0.613) -- (4.707,0.608) -- (5.053,0.597) -- (5.396,0.599) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.435,0.777) -- (5.138,0.963) -- (4.839,1.153) -- (4.540,1.340) -- (4.255,1.541) -- (3.945,1.713) -- (3.647,1.908) -- (3.357,2.101) -- (3.044,2.272) -- (2.748,2.459) -- (2.452,2.655) -- (2.151,2.838) -- (1.857,3.022) -- (1.566,3.223) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.255,0.786) -- (1.263,0.692) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.354,0.606) -- (1.446,0.593) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.351,0.880) -- (1.447,0.875) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.545,0.787) -- (1.534,0.694) -- (1.540,0.600);
  \draw[black] (1.607,3.183) -- (1.579,3.161) -- (1.560,3.133) -- (1.540,3.105) -- (1.514,3.082) -- (1.481,3.070) -- (1.453,3.051) -- (1.425,3.031) -- (1.392,3.021) -- (1.361,3.007) -- (1.327,3.006) -- (1.293,3.002) -- (1.260,2.992);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,3.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {9,9 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 (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LCS}\)) =  \(\dfrac{LS}{CS}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{9{,}9}{CS}\)
\par\addvspace{.5em}
 On a donc CS = \(\dfrac{9{,}9}{\sin(58^{\circ})}\) \(\approx\) 11,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.875) -- (1.044,2.655) -- (1.048,2.436) -- (1.051,2.221) -- (1.063,1.999) -- (1.062,1.783) -- (1.052,1.560) -- (1.062,1.343) -- (1.050,1.125);
  \draw[black] (1.050,1.125) -- (1.399,1.118) -- (1.749,1.121) -- (2.098,1.132) -- (2.451,1.128) -- (2.800,1.112) -- (3.147,1.122) -- (3.502,1.122) -- (3.850,1.123) -- (4.198,1.125) -- (4.549,1.127) -- (4.900,1.137) -- (5.248,1.120) -- (5.598,1.119) -- (5.950,1.125);
  \draw[black] (5.950,1.125) -- (5.619,1.233) -- (5.302,1.367) -- (4.969,1.465) -- (4.641,1.594) -- (4.317,1.713) -- (3.988,1.825) -- (3.661,1.937) -- (3.337,2.060) -- (3.006,2.165) -- (2.683,2.281) -- (2.359,2.411) -- (2.032,2.529) -- (1.703,2.645) -- (1.372,2.747) -- (1.050,2.875);
  \draw[black] (1.050,1.300) -- (1.053,1.240) -- (1.052,1.185) -- (1.050,1.125);
  \draw[black] (1.050,1.125) -- (1.109,1.128) -- (1.167,1.128) -- (1.225,1.125);
  \draw[black] (1.050,1.300) -- (1.110,1.306) -- (1.168,1.300) -- (1.225,1.300);
  \draw[black] (1.225,1.300) -- (1.220,1.242) -- (1.227,1.183) -- (1.225,1.125);
  \draw[black] (5.569,1.261) -- (5.573,1.248) -- (5.559,1.240) -- (5.570,1.225) -- (5.566,1.214) -- (5.564,1.203) -- (5.553,1.194) -- (5.560,1.181) -- (5.555,1.170) -- (5.545,1.160) -- (5.559,1.147) -- (5.554,1.136) -- (5.550,1.125);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.992) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.942) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.992) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.792) {4,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.172) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HDJ}\)) = \(\dfrac{HJ}{HD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HDJ}\)) =  \(\dfrac{1{,}5}{4{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HDJ}\) =  ArcTan(\(\dfrac{1{,}5}{4{,}2}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.990) -- (1.056,2.745) -- (1.059,2.492) -- (1.045,2.245) -- (1.053,1.998) -- (1.047,1.753) -- (1.050,1.504) -- (1.063,1.261) -- (1.050,1.010);
  \draw[black] (1.050,1.010) -- (1.398,1.001) -- (1.751,0.999) -- (2.100,1.005) -- (2.452,1.022) -- (2.803,1.000) -- (3.153,1.021) -- (3.498,1.004) -- (3.850,1.004) -- (4.203,1.006) -- (4.548,1.004) -- (4.899,1.023) -- (5.253,1.017) -- (5.602,1.017) -- (5.950,1.010);
  \draw[black] (5.950,1.010) -- (5.621,1.140) -- (5.302,1.280) -- (4.971,1.408) -- (4.649,1.549) -- (4.314,1.658) -- (3.991,1.814) -- (3.669,1.941) -- (3.341,2.074) -- (3.008,2.187) -- (2.677,2.322) -- (2.360,2.472) -- (2.028,2.593) -- (1.703,2.724) -- (1.377,2.858) -- (1.050,2.990);
  \draw[black] (1.050,1.208) -- (1.047,1.143) -- (1.052,1.075) -- (1.050,1.010);
  \draw[black] (1.050,1.010) -- (1.117,1.011) -- (1.181,1.016) -- (1.248,1.010);
  \draw[black] (1.050,1.208) -- (1.115,1.207) -- (1.182,1.208) -- (1.248,1.208);
  \draw[black] (1.248,1.208) -- (1.254,1.143) -- (1.247,1.076) -- (1.248,1.010);
  \draw[black] (1.415,2.842) -- (1.402,2.805) -- (1.390,2.766) -- (1.354,2.743) -- (1.339,2.704) -- (1.308,2.679) -- (1.273,2.660) -- (1.243,2.633) -- (1.206,2.617) -- (1.170,2.598) -- (1.127,2.605) -- (1.089,2.598) -- (1.050,2.589);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.877) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.057) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.877) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.677) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {1,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.371) {68°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BTH}\)) =  \(\dfrac{BH}{TH}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{BH}{1{,}6}\)

On a donc BH = 1,6 \(\times\) sin(68\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.743) -- (1.056,2.561) -- (1.059,2.369) -- (1.061,2.184) -- (1.055,1.999) -- (1.045,1.815) -- (1.062,1.631) -- (1.061,1.445) -- (1.050,1.257);
  \draw[black] (1.050,1.257) -- (1.400,1.267) -- (1.753,1.252) -- (2.103,1.248) -- (2.452,1.267) -- (2.802,1.246) -- (3.152,1.264) -- (3.500,1.244) -- (3.849,1.269) -- (4.200,1.249) -- (4.552,1.258) -- (4.901,1.254) -- (5.250,1.265) -- (5.602,1.249) -- (5.950,1.257);
  \draw[black] (5.950,1.257) -- (5.624,1.367) -- (5.297,1.464) -- (4.968,1.559) -- (4.646,1.664) -- (4.319,1.764) -- (3.991,1.846) -- (3.661,1.939) -- (3.332,2.043) -- (3.009,2.140) -- (2.684,2.241) -- (2.355,2.352) -- (2.031,2.444) -- (1.703,2.542) -- (1.373,2.632) -- (1.050,2.743);
  \draw[black] (1.050,1.405) -- (1.055,1.355) -- (1.048,1.307) -- (1.050,1.257);
  \draw[black] (1.050,1.257) -- (1.101,1.262) -- (1.149,1.259) -- (1.199,1.257);
  \draw[black] (1.050,1.405) -- (1.099,1.402) -- (1.149,1.411) -- (1.199,1.405);
  \draw[black] (1.199,1.405) -- (1.198,1.355) -- (1.199,1.306) -- (1.199,1.257);
  \draw[black] (1.442,2.624) -- (1.417,2.588) -- (1.392,2.553) -- (1.373,2.516) -- (1.343,2.486) -- (1.320,2.451) -- (1.291,2.418) -- (1.251,2.402) -- (1.213,2.383) -- (1.175,2.366) -- (1.136,2.347) -- (1.092,2.353) -- (1.050,2.338);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.123) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.810) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.123) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.141) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{CWN}\)) = \(\dfrac{CW}{WN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CWN}\)) = \(\dfrac{2{,}7}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{CWN}\) =ArcCos(\(\dfrac{2{,}7}{9{,}3}\)) \(\approx\)73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.656) -- (1.045,2.495) -- (1.039,2.325) -- (1.059,2.162) -- (1.060,1.999) -- (1.061,1.836) -- (1.039,1.675) -- (1.049,1.505) -- (1.050,1.344);
  \draw[black] (1.050,1.344) -- (1.400,1.333) -- (1.751,1.350) -- (2.098,1.344) -- (2.451,1.333) -- (2.801,1.343) -- (3.148,1.338) -- (3.500,1.353) -- (3.847,1.331) -- (4.198,1.333) -- (4.550,1.348) -- (4.900,1.349) -- (5.253,1.333) -- (5.600,1.341) -- (5.950,1.344);
  \draw[black] (5.950,1.344) -- (5.618,1.419) -- (5.299,1.527) -- (4.966,1.603) -- (4.644,1.684) -- (4.317,1.794) -- (3.993,1.879) -- (3.663,1.949) -- (3.335,2.041) -- (3.013,2.140) -- (2.682,2.206) -- (2.354,2.310) -- (2.035,2.401) -- (1.701,2.479) -- (1.373,2.558) -- (1.050,2.656);
  \draw[black] (1.050,1.475) -- (1.056,1.432) -- (1.049,1.388) -- (1.050,1.344);
  \draw[black] (1.050,1.344) -- (1.095,1.350) -- (1.136,1.346) -- (1.181,1.344);
  \draw[black] (1.050,1.475) -- (1.094,1.468) -- (1.139,1.475) -- (1.181,1.475);
  \draw[black] (1.181,1.475) -- (1.187,1.430) -- (1.183,1.387) -- (1.181,1.344);
  \draw[black] (5.572,1.445) -- (5.552,1.441) -- (5.552,1.432) -- (5.552,1.423) -- (5.562,1.412) -- (5.547,1.406) -- (5.558,1.395) -- (5.554,1.387) -- (5.559,1.378) -- (5.543,1.370) -- (5.552,1.361) -- (5.550,1.352) -- (5.549,1.344);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.210) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.723) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.210) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.010) {2,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NJT}\)) = \(\dfrac{NT}{NJ}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{NT}{2{,}6}\)
\par\addvspace{.5em}
On a donc NT =  2,6 \(\times\) tan(15\textdegree{})  \(\approx\) 0,7 cm\end{document}
