\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 110}

\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=110}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0110}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle CZD rectangle en C, on sait que :
\begin{itemize}
\item ZD=3,9 cm
\item \(\widehat{CZD}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle SAK rectangle en S, on sait que :
\begin{itemize}
\item SK=4,5 cm
\item AK=7,7 cm
\end{itemize}

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

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

 Dans le triangle TVN rectangle en T, on sait que :
\begin{itemize}
\item TV=7,9 cm
\item \(\widehat{TVN}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle WKS rectangle en W, on sait que :
\begin{itemize}
\item WK=3,1 cm
\item WS=5,2 cm
\end{itemize}

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

 Dans le triangle ZVJ rectangle en Z, on sait que :
\begin{itemize}
\item ZV=2,9 cm
\item \(\widehat{VJZ}\) =  17\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.428,3.053) -- (1.427,2.700) -- (1.414,2.352) -- (1.414,1.999) -- (1.434,1.652) -- (1.425,1.302) -- (1.432,0.953) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.769,0.604) -- (2.114,0.605) -- (2.461,0.592) -- (2.809,0.609) -- (3.151,0.594) -- (3.498,0.603) -- (3.845,0.600) -- (4.195,0.594) -- (4.537,0.604) -- (4.884,0.603) -- (5.229,0.612) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.295,0.776) -- (5.020,0.976) -- (4.740,1.153) -- (4.464,1.335) -- (4.195,1.541) -- (3.919,1.722) -- (3.638,1.912) -- (3.367,2.106) -- (3.083,2.282) -- (2.807,2.468) -- (2.523,2.643) -- (2.257,2.846) -- (1.975,3.023) -- (1.692,3.206) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.431,0.788) -- (1.425,0.692) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.519,0.604) -- (1.612,0.601) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.518,0.883) -- (1.612,0.874) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.709,0.788) -- (1.703,0.693) -- (1.704,0.600);
  \draw[black] (1.756,3.176) -- (1.738,3.149) -- (1.715,3.126) -- (1.698,3.096) -- (1.664,3.086) -- (1.640,3.063) -- (1.612,3.048) -- (1.585,3.027) -- (1.549,3.030) -- (1.519,3.019) -- (1.489,3.007) -- (1.457,3.000) -- (1.424,3.001);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,3.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.924,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {3,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{CZD}\)) = \(\dfrac{CZ}{ZD}\)
\par\addvspace{.5em}
d'où cos(56\textdegree{}) = \(\dfrac{CZ}{3{,}9}\)

On a donc CZ = 3,9 \(\times\) cos(56\textdegree{}) \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.492,3.400) -- (2.496,3.047) -- (2.490,2.698) -- (2.499,2.351) -- (2.485,1.998) -- (2.486,1.652) -- (2.493,1.300) -- (2.495,0.949) -- (2.492,0.600);
  \draw[black] (2.492,0.600) -- (2.742,0.606) -- (2.998,0.603) -- (3.246,0.587) -- (3.499,0.607) -- (3.750,0.606) -- (4.003,0.599) -- (4.255,0.612) -- (4.508,0.600);
  \draw[black] (4.508,0.600) -- (4.300,0.872) -- (4.097,1.153) -- (3.895,1.430) -- (3.704,1.723) -- (3.504,2.005) -- (3.292,2.273) -- (3.096,2.561) -- (2.900,2.841) -- (2.700,3.122) -- (2.492,3.400);
  \draw[black] (2.492,0.802) -- (2.497,0.733) -- (2.492,0.666) -- (2.492,0.600);
  \draw[black] (2.492,0.600) -- (2.557,0.605) -- (2.626,0.600) -- (2.693,0.600);
  \draw[black] (2.492,0.802) -- (2.559,0.797) -- (2.627,0.801) -- (2.693,0.802);
  \draw[black] (2.693,0.802) -- (2.689,0.734) -- (2.693,0.669) -- (2.693,0.600);
  \draw[black] (4.280,0.917) -- (4.256,0.898) -- (4.221,0.889) -- (4.208,0.858) -- (4.192,0.831) -- (4.161,0.814) -- (4.157,0.780) -- (4.141,0.753) -- (4.126,0.725) -- (4.121,0.693) -- (4.113,0.663) -- (4.108,0.632) -- (4.103,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.225,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.225,3.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.608,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,5 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 (3.915,0.837) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SKA}\)) = \(\dfrac{SK}{AK}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{SKA}\)) =  \(\dfrac{4{,}5}{7{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SKA}\) = Arccos (\(\dfrac{4{,}5}{7{,}7}\)) \(\approx\) 54\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.436,3.053) -- (1.425,2.702) -- (1.420,2.351) -- (1.412,2.003) -- (1.421,1.650) -- (1.418,1.303) -- (1.433,0.947) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.769,0.601) -- (2.117,0.611) -- (2.463,0.588) -- (2.809,0.588) -- (3.155,0.613) -- (3.503,0.598) -- (3.847,0.592) -- (4.191,0.610) -- (4.538,0.593) -- (4.881,0.598) -- (5.227,0.599) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.290,0.777) -- (5.022,0.970) -- (4.747,1.158) -- (4.468,1.351) -- (4.194,1.541) -- (3.915,1.726) -- (3.639,1.903) -- (3.360,2.084) -- (3.083,2.272) -- (2.802,2.460) -- (2.530,2.650) -- (2.247,2.833) -- (1.972,3.023) -- (1.698,3.203) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.420,0.787) -- (1.420,0.693) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.517,0.597) -- (1.610,0.598) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.519,0.875) -- (1.610,0.878) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.708,0.787) -- (1.709,0.692) -- (1.704,0.600);
  \draw[black] (1.763,3.172) -- (1.736,3.150) -- (1.712,3.129) -- (1.692,3.103) -- (1.673,3.074) -- (1.640,3.064) -- (1.614,3.043) -- (1.582,3.034) -- (1.550,3.028) -- (1.520,3.016) -- (1.489,3.007) -- (1.457,2.998) -- (1.424,3.002);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,3.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.924,1.933) {7,9 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 TVN rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TVN}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{TVN}\)) = \(\dfrac{TN}{TV}\)
\par\addvspace{.5em}
d'où tan(56\textdegree{}) =\(\dfrac{TN}{7{,}9}\)
\par\addvspace{.5em}
On a donc TN = 7,9 \(\times\) tan(56\textdegree{}) \(\approx\) 11,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.152,3.400) -- (1.147,3.047) -- (1.154,2.697) -- (1.159,2.351) -- (1.154,2.003) -- (1.162,1.647) -- (1.143,1.302) -- (1.151,0.951) -- (1.152,0.600);
  \draw[black] (1.152,0.600) -- (1.487,0.598) -- (1.825,0.600) -- (2.161,0.598) -- (2.491,0.600) -- (2.831,0.598) -- (3.163,0.588) -- (3.500,0.603) -- (3.835,0.602) -- (4.172,0.600) -- (4.506,0.589) -- (4.839,0.613) -- (5.175,0.596) -- (5.512,0.595) -- (5.848,0.600);
  \draw[black] (5.848,0.600) -- (5.557,0.773) -- (5.256,0.942) -- (4.971,1.136) -- (4.672,1.303) -- (4.384,1.486) -- (4.089,1.650) -- (3.794,1.831) -- (3.496,1.996) -- (3.205,2.176) -- (2.912,2.351) -- (2.614,2.518) -- (2.332,2.710) -- (2.038,2.881) -- (1.738,3.047) -- (1.450,3.237) -- (1.152,3.400);
  \draw[black] (1.152,0.880) -- (1.153,0.786) -- (1.157,0.695) -- (1.152,0.600);
  \draw[black] (1.152,0.600) -- (1.245,0.606) -- (1.338,0.604) -- (1.432,0.600);
  \draw[black] (1.152,0.880) -- (1.244,0.883) -- (1.338,0.885) -- (1.432,0.880);
  \draw[black] (1.432,0.880) -- (1.430,0.787) -- (1.437,0.694) -- (1.432,0.600);
  \draw[black] (1.499,3.193) -- (1.484,3.161) -- (1.449,3.145) -- (1.426,3.120) -- (1.411,3.085) -- (1.383,3.064) -- (1.351,3.049) -- (1.322,3.030) -- (1.285,3.027) -- (1.253,3.016) -- (1.219,3.011) -- (1.187,2.992) -- (1.152,3.007);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.885,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.885,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.948,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.652,1.933) {3,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.481,2.754) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WKS}\)) = \(\dfrac{WS}{WK}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WKS}\)) = \(\dfrac{5{,}2}{3{,}1}\)
\par\addvspace{.5em}
 On a donc WKS = ArcTan(\(\dfrac{5{,}2}{3{,}1}\)) \(\approx\) 59\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.749) -- (1.041,2.560) -- (1.058,2.372) -- (1.037,2.189) -- (1.044,2.000) -- (1.052,1.810) -- (1.059,1.627) -- (1.058,1.440) -- (1.050,1.251);
  \draw[black] (1.050,1.251) -- (1.398,1.239) -- (1.751,1.248) -- (2.102,1.241) -- (2.453,1.249) -- (2.802,1.246) -- (3.149,1.264) -- (3.498,1.260) -- (3.850,1.241) -- (4.197,1.248) -- (4.549,1.248) -- (4.900,1.239) -- (5.247,1.255) -- (5.602,1.254) -- (5.950,1.251);
  \draw[black] (5.950,1.251) -- (5.619,1.345) -- (5.302,1.457) -- (4.968,1.542) -- (4.640,1.645) -- (4.315,1.738) -- (3.991,1.858) -- (3.664,1.955) -- (3.337,2.060) -- (3.010,2.157) -- (2.685,2.259) -- (2.357,2.358) -- (2.031,2.459) -- (1.701,2.552) -- (1.379,2.647) -- (1.050,2.749);
  \draw[black] (1.050,1.401) -- (1.052,1.350) -- (1.053,1.301) -- (1.050,1.251);
  \draw[black] (1.050,1.251) -- (1.099,1.255) -- (1.150,1.251) -- (1.200,1.251);
  \draw[black] (1.050,1.401) -- (1.100,1.404) -- (1.149,1.401) -- (1.200,1.401);
  \draw[black] (1.200,1.401) -- (1.206,1.350) -- (1.197,1.302) -- (1.200,1.251);
  \draw[black] (5.573,1.366) -- (5.570,1.357) -- (5.554,1.351) -- (5.566,1.338) -- (5.560,1.329) -- (5.557,1.320) -- (5.560,1.309) -- (5.561,1.299) -- (5.553,1.290) -- (5.551,1.281) -- (5.545,1.271) -- (5.552,1.261) -- (5.554,1.251);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.118) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.816) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.118) {J};
  \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 (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.283) {17°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZJV}\)) = \(\dfrac{ZV}{VJ}\)
\par\addvspace{.5em}
 D'où sin(17\textdegree{}) = \(\dfrac{2{,}9}{VJ}\)
\par\addvspace{.5em}
On a donc VJ =\(\dfrac{2{,}9}{\sin(17^{\circ})}\) \(\approx\) 9,9 cm\end{document}
