\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 227}

\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=227}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0227}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle VJT rectangle en V, on sait que :
\begin{itemize}
\item VT=6,2 cm
\item \(\widehat{VJT}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle VFR rectangle en V, on sait que :
\begin{itemize}
\item VF=0,5 cm
\item \(\widehat{VFR}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZRG rectangle en Z, on sait que :
\begin{itemize}
\item ZR=2,1 cm
\item ZG=3,6 cm
\end{itemize}

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

 Dans le triangle SWL rectangle en S, on sait que :
\begin{itemize}
\item SW=4,1 cm
\item \(\widehat{WLS}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSF rectangle en K, on sait que :
\begin{itemize}
\item KF=4,6 cm
\item SF=8,4 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.940) -- (1.048,2.707) -- (1.062,2.470) -- (1.063,2.234) -- (1.060,2.003) -- (1.038,1.765) -- (1.057,1.528) -- (1.061,1.292) -- (1.050,1.060);
  \draw[black] (1.050,1.060) -- (1.402,1.071) -- (1.747,1.058) -- (2.098,1.051) -- (2.453,1.060) -- (2.800,1.059) -- (3.149,1.062) -- (3.498,1.047) -- (3.847,1.048) -- (4.200,1.047) -- (4.550,1.051) -- (4.903,1.058) -- (5.251,1.066) -- (5.601,1.064) -- (5.950,1.060);
  \draw[black] (5.950,1.060) -- (5.618,1.175) -- (5.293,1.304) -- (4.975,1.443) -- (4.647,1.562) -- (4.320,1.697) -- (3.991,1.807) -- (3.667,1.942) -- (3.343,2.070) -- (3.007,2.188) -- (2.684,2.313) -- (2.355,2.434) -- (2.030,2.555) -- (1.703,2.695) -- (1.372,2.811) -- (1.050,2.940);
  \draw[black] (1.050,1.248) -- (1.049,1.186) -- (1.050,1.121) -- (1.050,1.060);
  \draw[black] (1.050,1.060) -- (1.111,1.064) -- (1.174,1.064) -- (1.238,1.060);
  \draw[black] (1.050,1.248) -- (1.112,1.252) -- (1.176,1.246) -- (1.238,1.248);
  \draw[black] (1.238,1.248) -- (1.236,1.185) -- (1.242,1.123) -- (1.238,1.060);
  \draw[black] (1.417,2.800) -- (1.404,2.762) -- (1.382,2.729) -- (1.358,2.698) -- (1.342,2.658) -- (1.306,2.639) -- (1.280,2.606) -- (1.240,2.594) -- (1.207,2.570) -- (1.170,2.553) -- (1.128,2.557) -- (1.090,2.539) -- (1.050,2.533);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.926) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.007) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.926) {T};
  \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 (3.500,0.726) {6,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.428,2.324) {69°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VJT}\)) = \(\dfrac{VT}{VJ}\)
\par\addvspace{.5em}
tan(69\textdegree{}) = \(\dfrac{6{,}2}{VJ}\)
\par\addvspace{.5em}
 On a donc VJ = \(\dfrac{6{,}2}{\tan(69^{\circ})}\) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.703) -- (1.041,2.525) -- (1.047,2.349) -- (1.051,2.174) -- (1.042,2.002) -- (1.040,1.826) -- (1.044,1.651) -- (1.061,1.472) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.399,1.294) -- (1.749,1.302) -- (2.103,1.299) -- (2.449,1.293) -- (2.800,1.305) -- (3.152,1.288) -- (3.498,1.297) -- (3.852,1.300) -- (4.200,1.301) -- (4.552,1.294) -- (4.901,1.306) -- (5.252,1.292) -- (5.601,1.286) -- (5.950,1.297);
  \draw[black] (5.950,1.297) -- (5.625,1.403) -- (5.300,1.489) -- (4.968,1.573) -- (4.647,1.677) -- (4.311,1.758) -- (3.990,1.859) -- (3.665,1.954) -- (3.338,2.057) -- (3.010,2.132) -- (2.688,2.238) -- (2.352,2.315) -- (2.024,2.412) -- (1.705,2.512) -- (1.377,2.603) -- (1.050,2.703);
  \draw[black] (1.050,1.438) -- (1.043,1.392) -- (1.045,1.343) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.097,1.299) -- (1.145,1.297) -- (1.191,1.297);
  \draw[black] (1.050,1.438) -- (1.097,1.442) -- (1.144,1.443) -- (1.191,1.438);
  \draw[black] (1.191,1.438) -- (1.191,1.391) -- (1.194,1.345) -- (1.191,1.297);
  \draw[black] (1.444,2.590) -- (1.417,2.553) -- (1.403,2.512) -- (1.375,2.479) -- (1.358,2.438) -- (1.319,2.416) -- (1.288,2.387) -- (1.255,2.360) -- (1.221,2.330) -- (1.177,2.323) -- (1.135,2.315) -- (1.093,2.302) -- (1.050,2.301);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.164) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.769) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.164) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {0,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.964) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.451,2.103) {74°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VFR}\)) = \(\dfrac{VR}{VF}\)
\par\addvspace{.5em}
d'où tan(74\textdegree{}) =\(\dfrac{VR}{0{,}5}\)
\par\addvspace{.5em}
On a donc VR = 0,5 \(\times\) tan(74\textdegree{}) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.100,3.400) -- (1.100,3.048) -- (1.095,2.702) -- (1.092,2.349) -- (1.111,1.997) -- (1.088,1.650) -- (1.110,1.300) -- (1.113,0.951) -- (1.100,0.600);
  \draw[black] (1.100,0.600) -- (1.445,0.609) -- (1.788,0.587) -- (2.131,0.591) -- (2.469,0.588) -- (2.813,0.603) -- (3.155,0.587) -- (3.501,0.601) -- (3.841,0.604) -- (4.185,0.596) -- (4.528,0.606) -- (4.875,0.613) -- (5.217,0.596) -- (5.556,0.589) -- (5.900,0.600);
  \draw[black] (5.900,0.600) -- (5.599,0.774) -- (5.298,0.949) -- (5.005,1.129) -- (4.706,1.311) -- (4.394,1.468) -- (4.097,1.640) -- (3.797,1.825) -- (3.494,1.996) -- (3.195,2.168) -- (2.893,2.340) -- (2.603,2.529) -- (2.301,2.702) -- (2.000,2.879) -- (1.702,3.047) -- (1.393,3.218) -- (1.100,3.400);
  \draw[black] (1.100,0.880) -- (1.096,0.788) -- (1.106,0.694) -- (1.100,0.600);
  \draw[black] (1.100,0.600) -- (1.192,0.604) -- (1.286,0.603) -- (1.380,0.600);
  \draw[black] (1.100,0.880) -- (1.193,0.877) -- (1.286,0.874) -- (1.380,0.880);
  \draw[black] (1.380,0.880) -- (1.374,0.786) -- (1.379,0.692) -- (1.380,0.600);
  \draw[black] (1.443,3.200) -- (1.434,3.164) -- (1.399,3.148) -- (1.375,3.123) -- (1.360,3.089) -- (1.330,3.069) -- (1.304,3.045) -- (1.266,3.042) -- (1.237,3.022) -- (1.203,3.013) -- (1.170,3.003) -- (1.135,2.999) -- (1.100,3.005);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.833,0.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.833,3.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.000,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.600,1.933) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {3,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.432,2.755) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZRG}\)) = \(\dfrac{ZG}{ZR}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ZRG}\)) = \(\dfrac{3{,}6}{2{,}1}\)
\par\addvspace{.5em}
 On a donc ZRG = ArcTan(\(\dfrac{3{,}6}{2{,}1}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.708,3.400) -- (1.704,3.048) -- (1.711,2.699) -- (1.708,2.353) -- (1.716,2.000) -- (1.709,1.649) -- (1.713,1.301) -- (1.710,0.951) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (2.066,0.609) -- (2.425,0.602) -- (2.786,0.588) -- (3.141,0.611) -- (3.501,0.604) -- (3.859,0.606) -- (4.214,0.601) -- (4.572,0.596) -- (4.931,0.604) -- (5.292,0.600);
  \draw[black] (5.292,0.600) -- (5.010,0.809) -- (4.731,1.024) -- (4.462,1.238) -- (4.189,1.463) -- (3.921,1.684) -- (3.639,1.888) -- (3.356,2.100) -- (3.085,2.323) -- (2.803,2.532) -- (2.532,2.750) -- (2.259,2.965) -- (1.984,3.183) -- (1.708,3.400);
  \draw[black] (1.708,0.880) -- (1.706,0.787) -- (1.703,0.694) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (1.801,0.602) -- (1.895,0.603) -- (1.988,0.600);
  \draw[black] (1.708,0.880) -- (1.801,0.877) -- (1.896,0.883) -- (1.988,0.880);
  \draw[black] (1.988,0.880) -- (1.984,0.786) -- (1.991,0.692) -- (1.988,0.600);
  \draw[black] (4.975,0.847) -- (4.966,0.827) -- (4.956,0.807) -- (4.947,0.787) -- (4.927,0.773) -- (4.928,0.748) -- (4.905,0.733) -- (4.899,0.711) -- (4.898,0.688) -- (4.898,0.666) -- (4.902,0.643) -- (4.883,0.623) -- (4.894,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,3.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.392,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.208,1.933) {4,1 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 (4.662,0.750) {38°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SLW}\)) = \(\dfrac{SW}{WL}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{4{,}1}{WL}\)
\par\addvspace{.5em}
On a donc WL =\(\dfrac{4{,}1}{\sin(38^{\circ})}\) \(\approx\) 6,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.584,3.400) -- (2.597,3.048) -- (2.571,2.697) -- (2.584,2.352) -- (2.591,2.003) -- (2.578,1.653) -- (2.575,1.303) -- (2.574,0.953) -- (2.584,0.600);
  \draw[black] (2.584,0.600) -- (2.815,0.607) -- (3.043,0.599) -- (3.270,0.597) -- (3.501,0.587) -- (3.728,0.603) -- (3.957,0.612) -- (4.188,0.600) -- (4.416,0.600);
  \draw[black] (4.416,0.600) -- (4.235,0.881) -- (4.054,1.161) -- (3.860,1.436) -- (3.690,1.727) -- (3.502,1.998) -- (3.320,2.280) -- (3.137,2.558) -- (2.945,2.838) -- (2.770,3.122) -- (2.584,3.400);
  \draw[black] (2.584,0.783) -- (2.582,0.722) -- (2.587,0.663) -- (2.584,0.600);
  \draw[black] (2.584,0.600) -- (2.643,0.607) -- (2.707,0.601) -- (2.767,0.600);
  \draw[black] (2.584,0.783) -- (2.645,0.777) -- (2.707,0.777) -- (2.767,0.783);
  \draw[black] (2.767,0.783) -- (2.771,0.721) -- (2.768,0.663) -- (2.767,0.600);
  \draw[black] (4.200,0.931) -- (4.172,0.913) -- (4.149,0.890) -- (4.126,0.866) -- (4.106,0.842) -- (4.087,0.815) -- (4.059,0.793) -- (4.046,0.762) -- (4.036,0.730) -- (4.020,0.700) -- (4.028,0.665) -- (4.018,0.633) -- (4.025,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.317,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.317,3.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.516,0.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.830,0.850) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{KFS}\)) = \(\dfrac{KF}{SF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{KFS}\)) =  \(\dfrac{4{,}6}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{KFS}\) = Arccos (\(\dfrac{4{,}6}{8{,}4}\)) \(\approx\) 57\textdegree{} \end{document}
