\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 278}

\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=278}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0278}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle FVW rectangle en F, on sait que :
\begin{itemize}
\item FV=1,7 cm
\item VW=9,9 cm
\end{itemize}

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

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

 Dans le triangle RKN rectangle en R, on sait que :
\begin{itemize}
\item RK=4,6 cm
\item \(\widehat{RKN}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle GSF rectangle en G, on sait que :
\begin{itemize}
\item SF=8 cm
\item \(\widehat{SFG}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle JTM rectangle en J, on sait que :
\begin{itemize}
\item TM=1,1 cm
\item \(\widehat{JTM}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle GFK rectangle en G, on sait que :
\begin{itemize}
\item GF=1 cm
\item GK=3,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GFK}\). \newpage
\section*{\textcolor{red}{Correction}}

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.427) -- (1.037,2.323) -- (1.044,2.212) -- (1.037,2.107) -- (1.056,2.003) -- (1.041,1.891) -- (1.037,1.789) -- (1.051,1.682) -- (1.050,1.573);
  \draw[black] (1.050,1.573) -- (1.401,1.580) -- (1.751,1.586) -- (2.098,1.575) -- (2.452,1.563) -- (2.797,1.584) -- (3.148,1.580) -- (3.503,1.574) -- (3.849,1.580) -- (4.202,1.565) -- (4.548,1.561) -- (4.903,1.585) -- (5.252,1.576) -- (5.599,1.571) -- (5.950,1.573);
  \draw[black] (5.950,1.573) -- (5.600,1.643) -- (5.249,1.690) -- (4.899,1.758) -- (4.549,1.827) -- (4.201,1.873) -- (3.852,1.933) -- (3.504,2.007) -- (3.153,2.070) -- (2.802,2.121) -- (2.452,2.192) -- (2.102,2.249) -- (1.751,2.298) -- (1.397,2.358) -- (1.050,2.427);
  \draw[black] (1.050,1.658) -- (1.046,1.631) -- (1.051,1.603) -- (1.050,1.573);
  \draw[black] (1.050,1.573) -- (1.078,1.568) -- (1.108,1.569) -- (1.135,1.573);
  \draw[black] (1.050,1.658) -- (1.077,1.652) -- (1.106,1.652) -- (1.135,1.658);
  \draw[black] (1.135,1.658) -- (1.132,1.630) -- (1.136,1.602) -- (1.135,1.573);
  \draw[black] (5.563,1.640) -- (5.549,1.637) -- (5.545,1.632) -- (5.550,1.625) -- (5.557,1.618) -- (5.549,1.613) -- (5.560,1.607) -- (5.546,1.602) -- (5.552,1.596) -- (5.553,1.590) -- (5.558,1.584) -- (5.559,1.579) -- (5.558,1.573);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.440) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.494) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.440) {W};
  \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 (3.667,2.000) {9,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.564) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FWV}\)) = \(\dfrac{FV}{VW}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FWV}\)) =\(\dfrac{1{,}7}{9{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FWV}\) = ArcSin(\(\dfrac{1{,}7}{9{,}9}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.573,3.400) -- (1.567,3.050) -- (1.582,2.700) -- (1.570,2.352) -- (1.582,2.001) -- (1.580,1.650) -- (1.569,1.302) -- (1.565,0.949) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.921,0.603) -- (2.275,0.600) -- (2.622,0.606) -- (2.976,0.597) -- (3.325,0.591) -- (3.673,0.593) -- (4.026,0.609) -- (4.376,0.612) -- (4.726,0.593) -- (5.077,0.588) -- (5.427,0.600);
  \draw[black] (5.427,0.600) -- (5.154,0.799) -- (4.883,1.004) -- (4.599,1.199) -- (4.328,1.404) -- (4.054,1.607) -- (3.777,1.804) -- (3.500,1.996) -- (3.229,2.209) -- (2.954,2.400) -- (2.668,2.594) -- (2.393,2.794) -- (2.128,3.010) -- (1.845,3.191) -- (1.573,3.400);
  \draw[black] (1.573,0.880) -- (1.577,0.788) -- (1.575,0.692) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.668,0.604) -- (1.761,0.597) -- (1.853,0.600);
  \draw[black] (1.573,0.880) -- (1.668,0.884) -- (1.761,0.876) -- (1.853,0.880);
  \draw[black] (1.853,0.880) -- (1.849,0.785) -- (1.854,0.693) -- (1.853,0.600);
  \draw[black] (1.904,3.160) -- (1.878,3.140) -- (1.858,3.115) -- (1.837,3.091) -- (1.802,3.084) -- (1.783,3.058) -- (1.750,3.052) -- (1.729,3.024) -- (1.695,3.023) -- (1.665,3.015) -- (1.636,3.002) -- (1.605,2.993) -- (1.573,2.998);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.527,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.073,1.933) {4,6 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 (1.876,2.739) {54°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{RKN}\)) = \(\dfrac{RK}{KN}\)
\par\addvspace{.5em}
 d'où cos(54\textdegree{}) = \(\dfrac{4{,}6}{KN}\)
\par\addvspace{.5em}
On a donc KN = \(\dfrac{4{,}6}{\cos(54^{\circ})}\)\(\approx\) 7,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.432) -- (1.057,2.325) -- (1.040,2.218) -- (1.044,2.109) -- (1.060,1.999) -- (1.042,1.895) -- (1.043,1.782) -- (1.039,1.676) -- (1.050,1.568);
  \draw[black] (1.050,1.568) -- (1.400,1.572) -- (1.753,1.569) -- (2.102,1.557) -- (2.448,1.580) -- (2.800,1.564) -- (3.151,1.559) -- (3.501,1.560) -- (3.848,1.568) -- (4.198,1.558) -- (4.550,1.574) -- (4.901,1.580) -- (5.252,1.568) -- (5.600,1.580) -- (5.950,1.568);
  \draw[black] (5.950,1.568) -- (5.602,1.638) -- (5.255,1.704) -- (4.900,1.747) -- (4.549,1.827) -- (4.199,1.869) -- (3.849,1.948) -- (3.499,1.987) -- (3.146,2.056) -- (2.800,2.130) -- (2.445,2.174) -- (2.104,2.253) -- (1.752,2.321) -- (1.398,2.372) -- (1.050,2.432);
  \draw[black] (1.050,1.654) -- (1.051,1.625) -- (1.055,1.597) -- (1.050,1.568);
  \draw[black] (1.050,1.568) -- (1.079,1.570) -- (1.109,1.571) -- (1.136,1.568);
  \draw[black] (1.050,1.654) -- (1.080,1.652) -- (1.108,1.655) -- (1.136,1.654);
  \draw[black] (1.136,1.654) -- (1.134,1.626) -- (1.132,1.597) -- (1.136,1.568);
  \draw[black] (5.562,1.636) -- (5.551,1.632) -- (5.554,1.626) -- (5.545,1.621) -- (5.544,1.616) -- (5.561,1.608) -- (5.548,1.603) -- (5.550,1.597) -- (5.560,1.591) -- (5.551,1.585) -- (5.541,1.580) -- (5.542,1.574) -- (5.552,1.568);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.435) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.499) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.435) {F};
  \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 (3.667,2.000) {8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.559) {10°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GFS}\)) = \(\dfrac{GS}{SF}\)
\par\addvspace{.5em}
 D'où sin(10\textdegree{}) = \(\dfrac{GS}{8}\)

On a donc GS = 8 \(\times\) sin(10\textdegree{}) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.303) -- (1.057,2.975) -- (1.042,2.649) -- (1.041,2.326) -- (1.049,1.997) -- (1.052,1.672) -- (1.048,1.346) -- (1.058,1.020) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.402,0.691) -- (1.753,0.690) -- (2.097,0.689) -- (2.450,0.696) -- (2.803,0.689) -- (3.152,0.689) -- (3.500,0.689) -- (3.847,0.692) -- (4.202,0.708) -- (4.548,0.692) -- (4.900,0.694) -- (5.249,0.690) -- (5.601,0.693) -- (5.950,0.697);
  \draw[black] (5.950,0.697) -- (5.642,0.852) -- (5.341,1.026) -- (5.030,1.180) -- (4.733,1.359) -- (4.415,1.508) -- (4.110,1.668) -- (3.806,1.834) -- (3.494,1.990) -- (3.187,2.153) -- (2.881,2.320) -- (2.579,2.490) -- (2.277,2.659) -- (1.966,2.810) -- (1.665,2.985) -- (1.355,3.137) -- (1.050,3.303);
  \draw[black] (1.050,0.958) -- (1.052,0.870) -- (1.044,0.784) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.137,0.694) -- (1.224,0.694) -- (1.311,0.697);
  \draw[black] (1.050,0.958) -- (1.138,0.956) -- (1.224,0.954) -- (1.311,0.958);
  \draw[black] (1.311,0.958) -- (1.304,0.870) -- (1.308,0.784) -- (1.311,0.697);
  \draw[black] (1.401,3.116) -- (1.381,3.087) -- (1.370,3.050) -- (1.336,3.031) -- (1.318,2.998) -- (1.282,2.986) -- (1.258,2.957) -- (1.228,2.934) -- (1.194,2.920) -- (1.157,2.917) -- (1.122,2.908) -- (1.086,2.906) -- (1.050,2.899);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.564) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.369) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.564) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.364) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {1,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.393,2.665) {62°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JTM}\)) =  \(\dfrac{JM}{TM}\)
\par\addvspace{.5em}
 d'où sin(62\textdegree{}) = \(\dfrac{JM}{1{,}1}\)

On a donc JM = 1,1 \(\times\) sin(62\textdegree{}) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.681) -- (1.059,2.509) -- (1.045,2.339) -- (1.040,2.167) -- (1.053,1.997) -- (1.042,1.828) -- (1.040,1.657) -- (1.056,1.490) -- (1.050,1.319);
  \draw[black] (1.050,1.319) -- (1.397,1.309) -- (1.750,1.311) -- (2.102,1.328) -- (2.450,1.315) -- (2.803,1.317) -- (3.148,1.318) -- (3.499,1.310) -- (3.849,1.327) -- (4.197,1.308) -- (4.551,1.309) -- (4.899,1.309) -- (5.253,1.313) -- (5.601,1.325) -- (5.950,1.319);
  \draw[black] (5.950,1.319) -- (5.622,1.405) -- (5.294,1.495) -- (4.971,1.597) -- (4.643,1.682) -- (4.317,1.785) -- (3.988,1.855) -- (3.661,1.958) -- (3.338,2.057) -- (3.013,2.145) -- (2.677,2.216) -- (2.355,2.317) -- (2.032,2.421) -- (1.700,2.497) -- (1.377,2.601) -- (1.050,2.681);
  \draw[black] (1.050,1.456) -- (1.047,1.410) -- (1.050,1.365) -- (1.050,1.319);
  \draw[black] (1.050,1.319) -- (1.096,1.315) -- (1.140,1.320) -- (1.186,1.319);
  \draw[black] (1.050,1.456) -- (1.095,1.456) -- (1.139,1.454) -- (1.186,1.456);
  \draw[black] (1.186,1.456) -- (1.189,1.409) -- (1.187,1.364) -- (1.186,1.319);
  \draw[black] (1.443,2.572) -- (1.422,2.532) -- (1.410,2.490) -- (1.376,2.459) -- (1.353,2.423) -- (1.322,2.393) -- (1.287,2.369) -- (1.254,2.341) -- (1.220,2.312) -- (1.177,2.304) -- (1.136,2.288) -- (1.093,2.284) -- (1.050,2.290);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.186) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.747) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.186) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.986) {3,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.453,2.083) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GFK}\)) = \(\dfrac{GK}{GF}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{GFK}\)) = \(\dfrac{3{,}6}{1}\)
\par\addvspace{.5em}
 On a donc GFK = ArcTan(3,6) \(\approx\) 74\textdegree{}\end{document}
