\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 157}

\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=157}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0157}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle WHN rectangle en W, on sait que :
\begin{itemize}
\item WN=6,5 cm
\item HN=7,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{WHN}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle VGK rectangle en V, on sait que :
\begin{itemize}
\item VG=2,8 cm
\item GK=8,6 cm
\end{itemize}

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

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

 Dans le triangle CDJ rectangle en C, on sait que :
\begin{itemize}
\item CJ=0,7 cm
\item \(\widehat{DJC}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle HFJ rectangle en H, on sait que :
\begin{itemize}
\item HJ=8 cm
\item \(\widehat{FJH}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle HSW rectangle en H, on sait que :
\begin{itemize}
\item HS=4,9 cm
\item \(\widehat{HSW}\) = 56\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.296,3.400) -- (1.292,3.048) -- (1.300,2.701) -- (1.284,2.350) -- (1.293,2.002) -- (1.285,1.648) -- (1.285,1.298) -- (1.300,0.948) -- (1.296,0.600);
  \draw[black] (1.296,0.600) -- (1.635,0.599) -- (1.971,0.613) -- (2.316,0.594) -- (2.654,0.593) -- (2.989,0.594) -- (3.330,0.597) -- (3.671,0.589) -- (4.009,0.601) -- (4.346,0.609) -- (4.686,0.596) -- (5.023,0.602) -- (5.368,0.605) -- (5.704,0.600);
  \draw[black] (5.704,0.600) -- (5.413,0.793) -- (5.118,0.978) -- (4.827,1.164) -- (4.528,1.344) -- (4.232,1.532) -- (3.937,1.711) -- (3.653,1.915) -- (3.350,2.089) -- (3.060,2.276) -- (2.767,2.475) -- (2.475,2.661) -- (2.176,2.834) -- (1.883,3.029) -- (1.598,3.220) -- (1.296,3.400);
  \draw[black] (1.296,0.880) -- (1.293,0.785) -- (1.291,0.695) -- (1.296,0.600);
  \draw[black] (1.296,0.600) -- (1.390,0.605) -- (1.482,0.599) -- (1.576,0.600);
  \draw[black] (1.296,0.880) -- (1.389,0.882) -- (1.484,0.885) -- (1.576,0.880);
  \draw[black] (1.576,0.880) -- (1.575,0.785) -- (1.578,0.692) -- (1.576,0.600);
  \draw[black] (1.630,3.188) -- (1.616,3.157) -- (1.589,3.136) -- (1.568,3.110) -- (1.541,3.091) -- (1.520,3.063) -- (1.488,3.050) -- (1.461,3.028) -- (1.426,3.026) -- (1.395,3.012) -- (1.364,2.996) -- (1.329,3.005) -- (1.296,2.992);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.029,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.029,3.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.804,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {6,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.617,2.749) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WHN}\)) =  \(\dfrac{WN}{HN}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{WHN}\)) = \(\dfrac{6{,}5}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{WHN}\) =  ArcSin(\(\dfrac{6{,}5}{7{,}7}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.844) -- (1.056,2.633) -- (1.057,2.419) -- (1.062,2.213) -- (1.054,2.000) -- (1.045,1.786) -- (1.043,1.578) -- (1.051,1.366) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.398,1.162) -- (1.753,1.167) -- (2.099,1.147) -- (2.450,1.158) -- (2.800,1.165) -- (3.153,1.151) -- (3.500,1.149) -- (3.852,1.167) -- (4.201,1.147) -- (4.552,1.150) -- (4.900,1.167) -- (5.252,1.150) -- (5.603,1.155) -- (5.950,1.156);
  \draw[black] (5.950,1.156) -- (5.619,1.260) -- (5.295,1.373) -- (4.971,1.491) -- (4.638,1.597) -- (4.315,1.723) -- (3.990,1.831) -- (3.666,1.947) -- (3.338,2.066) -- (3.006,2.157) -- (2.688,2.294) -- (2.355,2.395) -- (2.024,2.496) -- (1.703,2.627) -- (1.379,2.744) -- (1.050,2.844);
  \draw[black] (1.050,1.325) -- (1.055,1.268) -- (1.053,1.214) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.106,1.163) -- (1.164,1.161) -- (1.219,1.156);
  \draw[black] (1.050,1.325) -- (1.106,1.324) -- (1.161,1.322) -- (1.219,1.325);
  \draw[black] (1.219,1.325) -- (1.212,1.270) -- (1.212,1.213) -- (1.219,1.156);
  \draw[black] (5.574,1.286) -- (5.566,1.277) -- (5.570,1.264) -- (5.571,1.253) -- (5.566,1.243) -- (5.552,1.234) -- (5.558,1.222) -- (5.544,1.213) -- (5.551,1.201) -- (5.559,1.189) -- (5.552,1.178) -- (5.544,1.168) -- (5.554,1.156);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.023) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.910) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.023) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.200) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VKG}\)) = \(\dfrac{VG}{GK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{VKG}\)) =\(\dfrac{2{,}8}{8{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VKG}\) = ArcSin(\(\dfrac{2{,}8}{8{,}6}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.892) -- (1.045,2.671) -- (1.044,2.443) -- (1.059,2.223) -- (1.037,2.000) -- (1.052,1.780) -- (1.052,1.555) -- (1.037,1.334) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.398,1.107) -- (1.753,1.122) -- (2.097,1.100) -- (2.447,1.101) -- (2.800,1.104) -- (3.152,1.104) -- (3.499,1.116) -- (3.850,1.116) -- (4.197,1.108) -- (4.551,1.099) -- (4.901,1.103) -- (5.252,1.113) -- (5.597,1.114) -- (5.950,1.108);
  \draw[black] (5.950,1.108) -- (5.619,1.215) -- (5.296,1.353) -- (4.972,1.471) -- (4.640,1.580) -- (4.316,1.706) -- (3.994,1.827) -- (3.666,1.949) -- (3.331,2.048) -- (3.011,2.183) -- (2.685,2.296) -- (2.354,2.411) -- (2.025,2.525) -- (1.704,2.666) -- (1.380,2.774) -- (1.050,2.892);
  \draw[black] (1.050,1.287) -- (1.053,1.229) -- (1.047,1.169) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.108,1.105) -- (1.169,1.106) -- (1.228,1.108);
  \draw[black] (1.050,1.287) -- (1.110,1.285) -- (1.169,1.292) -- (1.228,1.287);
  \draw[black] (1.228,1.287) -- (1.224,1.229) -- (1.228,1.167) -- (1.228,1.108);
  \draw[black] (5.575,1.245) -- (5.576,1.232) -- (5.573,1.221) -- (5.564,1.212) -- (5.558,1.201) -- (5.549,1.191) -- (5.566,1.176) -- (5.558,1.166) -- (5.558,1.154) -- (5.561,1.142) -- (5.545,1.132) -- (5.554,1.120) -- (5.557,1.108);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.975) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.958) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.975) {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 (3.500,0.775) {0,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.157) {20°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CJD}\)) = \(\dfrac{CD}{CJ}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{CD}{0{,}7}\)
\par\addvspace{.5em}
On a donc CD =  0,7 \(\times\) tan(20\textdegree{})  \(\approx\) 0,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.844) -- (1.062,2.632) -- (1.050,2.421) -- (1.056,2.212) -- (1.062,2.003) -- (1.038,1.787) -- (1.052,1.581) -- (1.044,1.370) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.401,1.157) -- (1.750,1.169) -- (2.103,1.165) -- (2.449,1.150) -- (2.800,1.166) -- (3.147,1.152) -- (3.502,1.168) -- (3.850,1.158) -- (4.200,1.151) -- (4.547,1.167) -- (4.899,1.167) -- (5.249,1.165) -- (5.598,1.144) -- (5.950,1.156);
  \draw[black] (5.950,1.156) -- (5.626,1.278) -- (5.293,1.374) -- (4.970,1.501) -- (4.637,1.598) -- (4.314,1.706) -- (3.994,1.836) -- (3.663,1.943) -- (3.333,2.046) -- (3.007,2.163) -- (2.685,2.288) -- (2.356,2.390) -- (2.034,2.515) -- (1.704,2.613) -- (1.382,2.740) -- (1.050,2.844);
  \draw[black] (1.050,1.325) -- (1.051,1.270) -- (1.056,1.214) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.107,1.152) -- (1.161,1.160) -- (1.219,1.156);
  \draw[black] (1.050,1.325) -- (1.105,1.329) -- (1.162,1.329) -- (1.219,1.325);
  \draw[black] (1.219,1.325) -- (1.218,1.267) -- (1.222,1.213) -- (1.219,1.156);
  \draw[black] (5.566,1.289) -- (5.569,1.276) -- (5.565,1.266) -- (5.565,1.254) -- (5.569,1.242) -- (5.566,1.232) -- (5.555,1.222) -- (5.551,1.212) -- (5.551,1.201) -- (5.554,1.189) -- (5.551,1.178) -- (5.555,1.167) -- (5.553,1.156);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.023) {H};
  \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) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.823) {8 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 (5.292,1.200) {19°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HJF}\)) = \(\dfrac{HJ}{FJ}\)
\par\addvspace{.5em}
  d'où    cos(19\textdegree{}) =  \(\dfrac{8}{FJ}\)
\par\addvspace{.5em}
 On a donc FJ =  \(\dfrac{8}{\cos(19^{\circ})}\) \(\approx\) 8,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.424,3.052) -- (1.431,2.697) -- (1.426,2.353) -- (1.424,2.000) -- (1.429,1.650) -- (1.432,1.302) -- (1.415,0.952) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.774,0.587) -- (2.117,0.606) -- (2.460,0.607) -- (2.805,0.613) -- (3.157,0.588) -- (3.500,0.600) -- (3.844,0.609) -- (4.191,0.612) -- (4.538,0.605) -- (4.886,0.588) -- (5.229,0.602) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.297,0.781) -- (5.018,0.966) -- (4.743,1.150) -- (4.472,1.358) -- (4.185,1.528) -- (3.919,1.730) -- (3.643,1.908) -- (3.369,2.105) -- (3.094,2.289) -- (2.803,2.464) -- (2.532,2.655) -- (2.258,2.846) -- (1.987,3.036) -- (1.702,3.213) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.426,0.785) -- (1.424,0.694) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.516,0.596) -- (1.612,0.601) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.516,0.877) -- (1.610,0.884) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.702,0.788) -- (1.705,0.693) -- (1.704,0.600);
  \draw[black] (1.757,3.176) -- (1.743,3.145) -- (1.711,3.129) -- (1.696,3.099) -- (1.663,3.087) -- (1.641,3.062) -- (1.608,3.054) -- (1.580,3.039) -- (1.556,3.012) -- (1.522,3.008) -- (1.490,3.000) -- (1.457,3.006) -- (1.424,2.995);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,3.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.924,1.933) {4,9 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 (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HSW}\)) = \(\dfrac{HW}{HS}\)
\par\addvspace{.5em}
d'où tan(56\textdegree{}) =\(\dfrac{HW}{4{,}9}\)
\par\addvspace{.5em}
On a donc HW = 4,9 \(\times\) tan(56\textdegree{}) \(\approx\) 7,3 cm\end{document}
