\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 277}

\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=277}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0277}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle DMA rectangle en D, on sait que :
\begin{itemize}
\item DA=6,2 cm
\item \(\widehat{MAD}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle HBV rectangle en H, on sait que :
\begin{itemize}
\item HB=2,9 cm
\item HV=6,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HBV}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle JFA rectangle en J, on sait que :
\begin{itemize}
\item JA=2,5 cm
\item \(\widehat{FAJ}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle NZL rectangle en N, on sait que :
\begin{itemize}
\item NZ=1,7 cm
\item NL=5,1 cm
\end{itemize}

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

\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle HGS rectangle en H, on sait que :
\begin{itemize}
\item HG=1,3 cm
\item \(\widehat{HGS}\) = 54\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.091) -- (1.058,2.817) -- (1.062,2.545) -- (1.045,2.276) -- (1.053,2.002) -- (1.053,1.726) -- (1.037,1.452) -- (1.053,1.184) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.397,0.921) -- (1.751,0.907) -- (2.101,0.918) -- (2.449,0.922) -- (2.797,0.914) -- (3.148,0.920) -- (3.502,0.917) -- (3.852,0.911) -- (4.199,0.898) -- (4.548,0.900) -- (4.903,0.908) -- (5.249,0.905) -- (5.598,0.913) -- (5.950,0.909);
  \draw[black] (5.950,0.909) -- (5.640,1.035) -- (5.340,1.187) -- (5.027,1.316) -- (4.730,1.466) -- (4.420,1.597) -- (4.115,1.735) -- (3.799,1.854) -- (3.496,1.991) -- (3.200,2.143) -- (2.887,2.266) -- (2.580,2.409) -- (2.276,2.549) -- (1.965,2.677) -- (1.661,2.818) -- (1.360,2.967) -- (1.050,3.091);
  \draw[black] (1.050,1.127) -- (1.047,1.054) -- (1.046,0.983) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.123,0.910) -- (1.195,0.913) -- (1.268,0.909);
  \draw[black] (1.050,1.127) -- (1.121,1.129) -- (1.197,1.123) -- (1.268,1.127);
  \draw[black] (1.268,1.127) -- (1.266,1.055) -- (1.268,0.981) -- (1.268,0.909);
  \draw[black] (5.578,1.075) -- (5.586,1.056) -- (5.582,1.043) -- (5.568,1.033) -- (5.558,1.022) -- (5.559,1.007) -- (5.555,0.993) -- (5.555,0.979) -- (5.561,0.964) -- (5.550,0.951) -- (5.542,0.938) -- (5.546,0.923) -- (5.548,0.909);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.776) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.157) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.776) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.576) {6,2 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 (5.298,0.981) {24°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DAM}\)) = \(\dfrac{DA}{MA}\)
\par\addvspace{.5em}
  d'où    cos(24\textdegree{}) =  \(\dfrac{6{,}2}{MA}\)
\par\addvspace{.5em}
 On a donc MA =  \(\dfrac{6{,}2}{\cos(24^{\circ})}\) \(\approx\) 6,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.077) -- (1.040,2.804) -- (1.060,2.537) -- (1.037,2.272) -- (1.055,2.003) -- (1.044,1.731) -- (1.061,1.463) -- (1.048,1.192) -- (1.050,0.923);
  \draw[black] (1.050,0.923) -- (1.397,0.925) -- (1.751,0.925) -- (2.102,0.923) -- (2.448,0.916) -- (2.801,0.934) -- (3.152,0.915) -- (3.498,0.932) -- (3.850,0.913) -- (4.202,0.923) -- (4.549,0.916) -- (4.899,0.926) -- (5.250,0.922) -- (5.598,0.911) -- (5.950,0.923);
  \draw[black] (5.950,0.923) -- (5.637,1.050) -- (5.334,1.188) -- (5.033,1.335) -- (4.721,1.454) -- (4.420,1.605) -- (4.111,1.731) -- (3.811,1.868) -- (3.494,1.990) -- (3.195,2.140) -- (2.883,2.265) -- (2.585,2.410) -- (2.276,2.535) -- (1.967,2.663) -- (1.665,2.817) -- (1.360,2.947) -- (1.050,3.077);
  \draw[black] (1.050,1.139) -- (1.053,1.067) -- (1.050,0.994) -- (1.050,0.923);
  \draw[black] (1.050,0.923) -- (1.120,0.930) -- (1.195,0.921) -- (1.265,0.923);
  \draw[black] (1.050,1.139) -- (1.123,1.133) -- (1.195,1.135) -- (1.265,1.139);
  \draw[black] (1.265,1.139) -- (1.260,1.066) -- (1.266,0.996) -- (1.265,0.923);
  \draw[black] (1.411,2.918) -- (1.405,2.878) -- (1.373,2.852) -- (1.359,2.815) -- (1.323,2.795) -- (1.298,2.767) -- (1.271,2.738) -- (1.239,2.716) -- (1.202,2.702) -- (1.165,2.690) -- (1.127,2.680) -- (1.089,2.676) -- (1.050,2.674);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.790) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.143) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.790) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.590) {6,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.414,2.452) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HBV}\)) = \(\dfrac{HV}{HB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HBV}\)) = \(\dfrac{6{,}6}{2{,}9}\)
\par\addvspace{.5em}
 On a donc HBV = ArcTan(\(\dfrac{6{,}6}{2{,}9}\)) \(\approx\) 66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.844) -- (1.056,2.632) -- (1.054,2.419) -- (1.060,2.210) -- (1.051,1.997) -- (1.059,1.786) -- (1.057,1.580) -- (1.039,1.367) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.403,1.151) -- (1.749,1.148) -- (2.102,1.157) -- (2.450,1.150) -- (2.802,1.146) -- (3.152,1.147) -- (3.500,1.160) -- (3.848,1.161) -- (4.199,1.143) -- (4.550,1.165) -- (4.898,1.163) -- (5.248,1.158) -- (5.597,1.169) -- (5.950,1.156);
  \draw[black] (5.950,1.156) -- (5.626,1.267) -- (5.294,1.369) -- (4.967,1.490) -- (4.642,1.611) -- (4.317,1.713) -- (3.991,1.829) -- (3.668,1.949) -- (3.332,2.050) -- (3.009,2.177) -- (2.683,2.277) -- (2.353,2.384) -- (2.033,2.513) -- (1.705,2.626) -- (1.374,2.723) -- (1.050,2.844);
  \draw[black] (1.050,1.325) -- (1.057,1.270) -- (1.049,1.214) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.106,1.161) -- (1.162,1.154) -- (1.219,1.156);
  \draw[black] (1.050,1.325) -- (1.106,1.319) -- (1.164,1.330) -- (1.219,1.325);
  \draw[black] (1.219,1.325) -- (1.212,1.269) -- (1.212,1.212) -- (1.219,1.156);
  \draw[black] (5.580,1.284) -- (5.571,1.275) -- (5.570,1.264) -- (5.560,1.255) -- (5.565,1.243) -- (5.553,1.234) -- (5.563,1.221) -- (5.555,1.211) -- (5.559,1.200) -- (5.547,1.190) -- (5.542,1.179) -- (5.555,1.167) -- (5.549,1.156);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.023) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.910) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.023) {A};
  \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 (3.500,0.823) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.200) {19°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JAF}\)) = \(\dfrac{JF}{JA}\)
\par\addvspace{.5em}
d'où  tan(19\textdegree{}) =  \(\dfrac{JF}{2{,}5}\)
\par\addvspace{.5em}
On a donc JF =  2,5 \(\times\) tan(19\textdegree{})  \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.817) -- (1.050,2.610) -- (1.046,2.410) -- (1.047,2.202) -- (1.058,2.001) -- (1.047,1.794) -- (1.055,1.594) -- (1.046,1.386) -- (1.050,1.183);
  \draw[black] (1.050,1.183) -- (1.397,1.187) -- (1.750,1.186) -- (2.101,1.178) -- (2.451,1.172) -- (2.801,1.191) -- (3.150,1.171) -- (3.501,1.171) -- (3.851,1.172) -- (4.199,1.187) -- (4.547,1.176) -- (4.900,1.194) -- (5.253,1.181) -- (5.601,1.185) -- (5.950,1.183);
  \draw[black] (5.950,1.183) -- (5.629,1.301) -- (5.299,1.401) -- (4.972,1.510) -- (4.648,1.622) -- (4.321,1.731) -- (3.986,1.824) -- (3.664,1.941) -- (3.343,2.066) -- (3.010,2.168) -- (2.679,2.264) -- (2.359,2.387) -- (2.031,2.494) -- (1.704,2.605) -- (1.373,2.708) -- (1.050,2.817);
  \draw[black] (1.050,1.347) -- (1.049,1.294) -- (1.048,1.238) -- (1.050,1.183);
  \draw[black] (1.050,1.183) -- (1.105,1.185) -- (1.160,1.181) -- (1.213,1.183);
  \draw[black] (1.050,1.347) -- (1.104,1.341) -- (1.158,1.352) -- (1.213,1.347);
  \draw[black] (1.213,1.347) -- (1.211,1.293) -- (1.210,1.236) -- (1.213,1.183);
  \draw[black] (5.572,1.309) -- (5.567,1.300) -- (5.555,1.292) -- (5.570,1.277) -- (5.559,1.268) -- (5.563,1.257) -- (5.561,1.247) -- (5.548,1.238) -- (5.544,1.227) -- (5.558,1.215) -- (5.559,1.204) -- (5.547,1.194) -- (5.558,1.183);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.050) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.883) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.050) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.850) {5,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.223) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NLZ}\)) = \(\dfrac{NZ}{NL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NLZ}\)) =  \(\dfrac{1{,}7}{5{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NLZ}\) =  ArcTan(\(\dfrac{1{,}7}{5{,}1}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.573,3.400) -- (1.574,3.047) -- (1.561,2.703) -- (1.568,2.349) -- (1.576,2.001) -- (1.575,1.653) -- (1.574,1.302) -- (1.576,0.948) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.926,0.604) -- (2.276,0.612) -- (2.624,0.607) -- (2.971,0.599) -- (3.325,0.595) -- (3.676,0.604) -- (4.024,0.600) -- (4.375,0.603) -- (4.727,0.594) -- (5.077,0.594) -- (5.427,0.600);
  \draw[black] (5.427,0.600) -- (5.150,0.802) -- (4.881,1.007) -- (4.601,1.199) -- (4.332,1.404) -- (4.056,1.602) -- (3.780,1.809) -- (3.504,2.011) -- (3.216,2.192) -- (2.945,2.398) -- (2.675,2.601) -- (2.404,2.805) -- (2.119,2.991) -- (1.855,3.210) -- (1.573,3.400);
  \draw[black] (1.573,0.880) -- (1.567,0.788) -- (1.569,0.694) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.668,0.595) -- (1.760,0.600) -- (1.853,0.600);
  \draw[black] (1.573,0.880) -- (1.665,0.877) -- (1.759,0.881) -- (1.853,0.880);
  \draw[black] (1.853,0.880) -- (1.851,0.787) -- (1.854,0.692) -- (1.853,0.600);
  \draw[black] (1.892,3.168) -- (1.878,3.139) -- (1.851,3.122) -- (1.838,3.090) -- (1.805,3.081) -- (1.784,3.056) -- (1.755,3.042) -- (1.727,3.028) -- (1.698,3.016) -- (1.667,3.007) -- (1.636,3.002) -- (1.604,3.005) -- (1.573,3.008);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,3.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.527,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.073,1.933) {1,3 cm};
  \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 (1.876,2.739) {54°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HGS}\)) = \(\dfrac{HS}{HG}\)
\par\addvspace{.5em}
d'où tan(54\textdegree{}) =\(\dfrac{HS}{1{,}3}\)
\par\addvspace{.5em}
On a donc HS = 1,3 \(\times\) tan(54\textdegree{}) \(\approx\) 1,8 cm\end{document}
